Bank Math

2 men can do a work in x days
1 men can do a work in (2 × x) days
y women can do a work in 3 days
1 women can do a work in 3y days
1 man : 1 woman
Days - 2x : 3y
Efficiency - 3y : 2x

Lopa's wages : Rasel's wages
= Lopa's 1 day's work : Rasel's 1 day's work
= 1/3 : 1/2
= 2 : 3
∴ Lopa's share = Tk.(2/5) × 150
= Tk 60.

Question: A and B are two fixed points 5 cm apart and C is a point on AB such that AC is 3cm. if the length of AC is increased by 6%, the length of CB is decreased by-

Solution:
Given that,
AB = 5 cm (fixed)
Initially, AC = 3 cm
So, CB = AB - AC = 5 - 3 = 2 cm

Now,
Increase in AC = 6%
∴ Increase in AC = (106/100) × 3 = 3.18cm
And,
Decrease in CB = 5 - 3.18 = 1.82 cm
∴ Decrease = 2 - 1.82 = 0.18 cm

So Percentage of decrease = (0.18/2) × 100%
= (18/2)%
= 9%

Thus, the length of CB is decreased by 9%.

Question: In a class of 40 students, the average marks in an exam was 60. The average marks of the students who passed is 65 and the average marks of the students who failed is 40. How many students failed in the exam?

Solution:
Let the number of students who failed = x
Then, the number of students who passed = 40 - x

According to the question,
65(40 - x) + 40x = 40 × 60
⇒ 2600 - 65x + 40x = 2400
⇒ -25x = 2400 - 2600
⇒ -25x = -200
∴ x = 200/25
∴ x = 8

∴ Number of students who failed = 8.

w + x + x + y + y + w = - 4 + 25 + 15
⇒ 2 (w + x + y) = 36
⇒ w + x + y = 18
So, average of w, x, y = 18/3 = 6

Question: A machine produces 240 toys in 4/3 hours. How many toys can it produce in 25 minutes?

Solution:
Given,
Time = 4/3 hours = (4/3) × 60 = 80 minutes

In 80 minutes, the number of toys produced = 240
In 1 minute, the number of toys produced = 240/80 = 3
In 25 minutes, the number of toys produced  = 3 × 25 = 75

∴ The machine can produce 75 toys in 25 minutes.

Question: Find slope of the line perpendicular to the line y = (1/3)x - 7.

Solution: 
Given line, y = (1/3)x - 7
The slope of this line is m₁ = 1/3 ; [Comparing with  y = mx + c]

We know, 
If two lines are perpendicular, their slopes satisfy m₁⋅m₂ = - 1
Let m₂ be the slope of the perpendicular line. Then we get,
⇒ (1/3)⋅m2 = - 1
⇒ m₂ = - 1 × 3
∴ m₂ = - 3

So the slope of the line perpendicular to the given line is - 3.

 Question: {(1 - sin²45°)/(1 + sin²45°)} + tan²45° = ?

Solution:
Given that, 
 {(1 - sin²45°)/(1 + sin²45°)} + tan²45°
= {1 - (1/√2)²}/{1 + (1/√2)²} + (1)²   [∴ sin 45° = 1/√2 ও tan 45° = 1] 
= {1 - (1/2)}/{1 + (1/2)} + 1 
= {(2 - 1)/2}/{(2 + 1)/2} + 1 
= (1/2)/(3/2) + 1 
= (1/3) + 1
= (1 + 3)/3
= 4/3

12 জন পুরুষ 24 দিনে করে 1 অংশ
∴ 12 জন পুরুষ 1 দিনে করে 1/24 অংশ
আবার, 12 জন মহিলা 12 দিনে করে 1 অংশ
∴ 12 জন মহিলা 1 দিনে করে 1/12 অংশ
তাহলে, 12 জন পুরুষ ও 12 জন মহিলা 1 দিনে করে
= (1/24 + 1/12) অংশ
= (1+2)/24 অংশ
= 3/24 = 1/8অংশ

নারী এবং পুরুষেরা একত্রে 1/8 অংশ করে 1 দিনে
∴ তারা একত্রে 1 বা সম্পূর্ণ অংশ করে 1×8 = 8 দিনে

Question: If x³ = 125, then x² + x = ?

Solution: 
Given that, 
x³ = 125
⇒ x³ = 5³
∴ x = 5

Now, 
x² + x = 5² + 5 = 25 + 5 = 30

So the value of x² + x is 30

Question: Three pipes A, B, and C can fill a tank from empty to full in 40 minutes, 20 minutes, and 10 minutes respectively. When the tank is empty, all three pipes are opened. A, B, and C discharge chemical solutions P, Q, and R respectively. What is the proportion of the solution R in the liquid in the tank after 2 minutes?

Solution: 
Part filled by (A + B + C) in 2 minutes
= 2 [(1/40) + (1/20) + (1/10)]
= 2 × (7/40)
= 7/20

Part filled by C (solution R) in 2 minutes = 2/10
= 1/5

∴ Proportion of solution R = (1/5) × (20/7)
= 4/7

Let
Marked price = X
Cost Price(C.P.) = 1440
Sale price(S.P.) = 1440 + 25% of 1440
= Tk. 1800

S.P. = M.P. - 25% of M.P.
S.P. = X - 25% of X
S.P. = X - 0.25X
1800 = 0.75X
X = 1800/.25
= 2400
M.P. = Tk. 2400

1 2 3 4 5 6 (Shahadat) 7 (Nitu) 8 (Altaf) 9 10 11 12 13 14
Here, Althaf is 8th from front, Shankar is 9th from rear end and Nitu is between them
So minimum no. of boys standing in the queue = 14.

Question: If a³ - b³ = 117 and a - b = 3 What is the value of ab?

Solution:
Given,
a³ - b³ = 117
a - b = 3

We know,
⇒ (a - b)³ + 3ab(a - b) = a³ - b³
⇒ 27 + 3ab(3) = 117
⇒ 27 + 9ab = 117
⇒ 9ab = 117 - 27
⇒ 9ab = 90
⇒ ab = 90/9
∴ ab = 10

Question: If θ = 60°, then what is the value of (1 - tan²θ)/(1 + tan²θ)?

Solution: 
Given that, 
θ = 60°

Now, 
(1 - tan²θ)/(1 + tan²θ)
= {1 - (tan60°)²}/{1 + (tan60°)²}
= {1 - (√3)²}/{1 + (√3)²}
= (1 - 3)/(1 + 3)
= (- 2)/4
= - 1/2

Question: 10 years ago, Person X was 6 years younger than Person Y. The sum of their present ages is 50. Find the present age of X.

Solution:
Let present age of X = a years  
Present age of Y = b years  

Given that,
a + b = 50 .........(1)
And, 
10 years ago, X was 6 years younger than Y  
⇒ (a - 10) = (b - 10) - 6  
 ⇒ a - 10 = b - 16  
 ∴ a = b - 6  

Now put x = y - 6 in equation (1) then we get,
⇒ (b - 6) + y = 50  
⇒ 2b - 6 = 50  
⇒ 2b = 56  
⇒ b = 56/2 = 28  
∴ b = 28

∴ Present age of X = b - 6 = 28 - 6 = 22 years.

Question: A cistern 8 m long and 6 m wide contains water up to a depth of 1 m 50 cm. Find the total area of the wet surface.

Solution: 
Here, l = 8 m , b = 6 m and h = 1 m 50 cm = 1.5  m

The water wets the bottom surface and the four vertical walls up to the water depth. The area of the bottom is (l × b) 
The area of the four walls is the perimeter of the (base × height) of the water which is, 2[(l + b) × h] = 2lh + 2bh

The total wet surface area is the sum of these areas.
∴ Area = lb + 2lh + 2bh
= (8 × 6) + 2(8 × 1.5) + 2(6 × 1.5)
= 48 + 24 + 18
= 90 sq. meter 

Distance = 92 km,
As they walk in opposite direction, their relative Speed = 5 + 6.5 = 11.5 km/h
So, Required time = Distance/Relative speed = 92/11.5 = 8 h

Question: Rina completes 60% or 3/5 of a work in 12 days. Then she calls in Sima, and together they finish the remaining work in 6 days. How long would Sima alone take to complete the whole work?

Solution:
Here,
Rina does 3/5 of the work in 12 days
in one day, she does = (3/5 ÷ 12)
= 3/60
= 1/20 part

In 6 days, she does 6/20 = 3/10 part
work remaining = 2/5

Together, they do 2/5 in 6 days
Together in one day = (2/5) ÷ 6
= 2/30
= 1/15

Rina one day = 1/20
Sima one day = 1/15 - 1/20
= (4-3)/60
= 1/60

∴ Sima alone will take 60 days to complete the whole work.

Let the lengths of the line segments be x and x+2 cm
then,
(x+2)²−x²=32
x²+4x+4−x²=32
4x=28
x=7cm

Let,
Maala’s age = 4A and
Kala’s age = 3A
Then,
4A + 3A = 28
A = 4
Maala’s age = 16 years
and Kala’s age = 12 years
Proportion of their ages after 8 is = (16 + 8) : (12 + 8)
= 24 : 20
= 6 : 5

Question: The sum of fifth and thirteenth term of an arithmetic progression is 28. What is the sum of the first seventeen terms of that progression?

Solution:
In an arithmetic progression.
Let first term = a
Common difference = d

We know,
a_n = a + (n - 1)d
∴ a₅ = a + 4d and a₁₃ = a + 12d

Given that, 
fifth term + thirteenth term = 28
⇒ a₅ + a₁₃ = 28
⇒ a + 4d + a + 12d = 28
⇒ 2a + 16d = 28
⇒ 2(a + 8d) = 28
∴ a + 8d = 14  ........(1)

We need the sum of the first 17 terms.
S₁₇ = (n/2) × [2a + (n - 1)d]
= (17/2) × [2a + 16d]
= 17/2 × 2(a + 8d)
= 17 × (a + 8d)
= 17 × 14
= 238

So the sum of the first seventeen terms is 238.

Question: The side of an equilateral triangle is 6m. What is the height of the triangle?

Solution:
Given,
The side of an equilateral triangle = 6m

We know,
Area of an equilateral triangle = (√3/4) × 6²
= (√3/4) × 36
= 9√3

Let,
the height of the triangle = h

We also know,
(1/2) × base × height = area
⇒ (1/2) × 6 × h = 9√3
⇒ h = 9√3/3
∴ h = 3√3

So, the height of the triangle = 3√3 m

Question: If a = 0.202 , then the value of

is:

Solution:

সঠিক উত্তর 1.202 হবে,
কারণ (+) যোগ চিহ্ন দিয়ে বের করা রাশির উত্তর নেই।

Question: In the figure AC and BC are radii of circles. The length of AB is 8. If AC = 4, what is BC? (BC is tangent to the circle with center A.)

Solution:
Since BC is tangent to circle with centre A
∴ BC is perpendicular to AC.
ΔABC is a right-angled triangle.

So,
BC = √(AB² - AC²)
= √(8² - 4²)
= √(64 - 16)
= √48
= √(16 × 3)
= 4√3

Let their income be x
and expenditure be y

A's income = 5x
B's income = 4x

A's expenditure = 3y
B's expenditure = 2y

Both saves 1600
Income - expenditure = saves

5x - 3y = 1600 .... (i)
4x - 2y = 1600 .... (ii)
Substract (ii) × 3 from (i) × 2

10x - 6y - ( 12x - 6y ) = 3200 - 4800
10x - 6y - 12x + 6y = - 1600
- 2x = - 1600

x = 800
Putting the value of x in (i)
5x - 3y = 1600
4000 - 3y = 1600
- 3y = - 2400
y = 800

∴ Income of a = 5x = 4000

Let the Market Price of the product is MP.
Let the Original Cost Price of the product is CP.
Selling Price (Discounted Price) = 100% of MP - 20% of MP = 80% of MP ......... (1)
Profit made by selling at discounted price = 20% of CP ........... (2)
Apply the formula:
Profit = Selling Price - Original Cost Price
⇒ 20% of CP = 80% of MP - 100% CP
⇒ MP = (120×CP)/80 = 3/2 × CP
Now if product is sold without any discount, then,
Profit = Selling Price (without discount) - Original Cost Price
= Market Price - Original Cost Price
= MP - CP
= 3/2 CP - CP
= 1/2 CP
= 50% of CP

Question: A bag contains 4 green balls and 3 yellow balls. If two balls are drawn without replacement, what is the probability that both are yellow?

Solution:
Total balls = 4 green + 3 yellow = 7 balls.
Probability that the first ball is yellow = 3/7

After removing 1 yellow ball, we have:
Remaining yellow balls = 2
Total remaining balls = 6
So, the probability that the second ball is yellow = 2/6 = 1/3

∴ Total probability (both yellow) = 3/7 × 1/3 = 1/7

Let 1 man's 1 day's work = x and 1 woman's 1 day's work = y
Then, 4x + 6y = 1/8 and 3x + 7y = 1/10
Solving the two equations, we get
x = 11/400, y = 1/400
∴ 1 woman's 1 day's work = 1/400
⇒ 10 women's 1 day's work = ((1/400)×10) = 1/40
Hence, 10 women will complete the work in 40 days

1000000 × 0.75 × 0.80 × 0.85=  5,10,000