ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Question: In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the rate in the remaining 40 overs to each the target of 282 runs?

Solution:
First 10 overs total run was = (3.2 × 10) = 32

Required run rate = (282 - 32)/40
= 250/40
= 6.25

Question: The difference between a number and its two-fifth is 510. What is the ten percent of that number?

Solution:
Let,
the number will be x.

ATQ,
x - (2x/5) = 510
⇒ (5x - 2x)/5 = 510
⇒ 3x/5 = 510
⇒ 3x = 510 × 5
⇒ x = 2550/3
x = 850

∴ The number of 10% = 850 × (10/100) = 85 

Question: A man deposits Tk. 5000 in a Bank at 10% interest rate compounded annually. At the end of the three year, the total amount including interest will become?

Solution: 
Given,
Principal, P = 5000 Tk.
Rate of interest, r = 10% = 10/100 = 1/10
Time, n = 3 years.

We know,
Compound Amount = P (1 + r)ⁿ
= 5000 × (1 + 1/10)³
= 5000 × (11/10)³
= 5000 × (11/10) × (11/10) × (11/10)
= 6655

Question: If x and y are negative, then which of the following statements is always true?

Solution:
If x < 0 and y < 0 then xy > 0.
So, whenever x and y are negative, then xy is positive.

Example: If x = - 1 and y = - 1 then xy = (- 1) × (- 1) = 1 > 0

Question: If (2p + 1/p) = 4, the value of (p³ + 1/8p³) is -

Solution:
Given,
(2p + 1/p) = 4
Or, (1/2) (2p + 1/p) = 4 × 1/2
Or, p + 1/2p = 2

Now, (p³ + 1/8p³) = (p)³ + (1/2p)³
= (p + 1/2p)³ - 3 . p . 1/2p . (p + 1/2p)
= (2)³ - 3 . (1/2) . 2
= 8 - 3
= 5

Question: Three numbers are in the ratio 3 : 4 : 6 and their products is 1944. The smallest number is - 

Solution:
Let the number be 3x, 4x, 6x.

ATQ,
3x × 4x × 6x = 1944
⇒ 72x³ = 1944
⇒ x³ = 1944/72
⇒ x³ = 27
∴ x = 3

So the smallest number = 3x = 3 × 3 = 9

Question: The angle of elevation of the sun, when the length of the shadow of a tree √3 times the height of the tree is:

Solution:

Let,
AB = height of tree
BC= Shadow of tree
angle of elevation = C
∴  BC = √3 AB

We know,
tanC = AB/BC
⇒ tanC = AB/√3AB
⇒ tanC = 1/√3
⇒ tanC = tan30°
∴ C = 30°

Question: The sum of the present ages of the father and daughter is 90 years. If five years ago the age of father was four times that of his daughter, find the present age of the daughter.

Solution: 
Let the age of daughter 5 years ago = x
The age of father 5 years ago = 4x

ATQ,
x + 5 + 4x + 5 = 90
⇒ 5x + 10 = 90
⇒ 5x = 80
∴ x = 16

∴ Daughter present age = 16 + 5 = 21 years.

Question: A piece of wire 91 cm long is bent in the form of an isosceles triangle. If the ratio of one of the equal sides to the base is 5 : 3, then what is the length of the base?

Solution:
Given,
Ratio of one of the equal sides to the base is 5 : 3
Therefore, the sides are 5x, 3x, 5x.

91 cm piece of wire is bent to form an isosceles triangle.
Thus perimeter of triangle is 91 cm.

ATQ,
∴ 13x = 91
⇒ x = 7

Thus the length of the base = 3 × 7 = 21 cm.

Question: A train 300 metres long is running at a speed of 25 metre per second. It will cross a bridge of 200 metres long in:

Solution:
Speed of train = 25 m/s.
Total distance covered while passing bridge = 300 + 200 = 500 m

∴ Time = 500​/25 
= 20 seconds.

Question: A company pays rent of Tk. 20000 per month for office space to its owner. But if the company pays the annual rent at the beginning of the year the owner gives a discount of 5% on the total annual rent. What is the annual amount the company pays to the owner after the discount?

Solution:
Total annual rent = Tk. (20000 × 12) = Tk. 240000
Discount = 5% of tk. 240000
= Tk. 12000
∴ Annual rent paid after discount = Tk. (240000 - 12000)
= Tk. 228000

Question: A train passes two bridges of length 800 m and 400 m in 100 seconds and 60 seconds respectively. The length of the train is -

Solution:
Let length of the train be x m and speed of the train is s kmph.
Speed, s = (x + 800)/100 . . . . . (i)
Speed, s = (x + 400)/60. . . . . (ii) 

Equating equation (i) and (ii),
we get,
(x + 800)/100 = (x + 400)/60
Or, (x + 800)/5 = (x + 400)/3
Or, 5x + 2000 = 3x + 2400
Or, 2x = 400
∴ x = 200m

∴ The length of the train is 200 meters.

Question: A man travels 50 km at speed 25 km/hr and next 40 km at 20 km/hr and there after travel 90 km at 15 km/hr. His average speed is:

Solution:
We know,
Average speed = Total distance/Total time
= (50 + 40 + 90)/(2 + 2 + 6) km/hr
= 180/10 km/hr
= 18 km/hr.

Question: Let O be the in-center of a triangle ABC and D be a point on the side BC of ΔABC, such that OD ⊥ BC. If ∠BOD = 60°, then ∠ABC = ?

Solution:

Given,
∠BOD = 60°

We know,
∴ ∠BDO + ∠DOB + ∠DBO = 180°
⇒ ∠DBO = 180° - ∠BDO + ∠DOB
⇒ ∠DBO = 180° - 90° - 60°
∴ ∠DBO = 30°

Now,
∠ABC = 2 × ∠DBO
⇒ ∠ABC = 2 × 30°
∴ ∠ABC = 60°

Question: The product of the roots of the equation 2a² - 5a + m = 10 is - 3. Find the value of m.

Solution:
Rearranging the given equation we have 2a² - 5a + (m - 10) = 0

We know that,
if ax² + bx + c = 0 is a quadratic equation, then the product of their roots = c/a

Given the product of the roots = - 3
⇒ (m - 10)/2 = - 3
⇒ (m - 10) = - 6
⇒ m = - 6 + 10
∴ m = 4

Question: A father is three times as old as his son. After fifteen years father will be twice as old as his son age that time. What is the present age of the son?

Solution:
Let, the son's age be x years
and the father's age is = 3x years

ATQ,
3x + 15 = 2(x + 15)
⇒ 3x + 15 = 2x + 30
∴ x = 15

∴ Son's age 15 years.

Question: A sum of money is sufficient to pay A's wages for 21 days or B's wages for 28 days. The same money is sufficient to pay the wages of both for?

Solution:
Let
total money be Tk. x
A's 1 day's wages = Tk. x/21
B's 1 day's wages = Tk. x/28

∴ (A + B)'s 1 day's wages = Tk. (x/21 + x/28)
= Tk. (4x + 3x)/84
= Tk. 7x/84
= Tk. x/12

∴ Money is sufficient to pay the wages of both for 12 days.

Question: A Zoo keeper counted the heads of the animals in a zoo and found it to be 80. When he counted the legs of the animals he found it to be 260. If the zoo had either pigeons or horses, how many horses were there in the zoo?

Solution: 
let, there are x horses and y pigeons.

ATQ,
x + y = 80 
⇒ x = 80 - y

and
4x + 2y = 260
⇒ 4 (80 - y) + 2y = 260 
⇒ 320 - 4y + 2y = 260 
⇒ 320 - 2y = 260
⇒ - 2y = 260 - 320
⇒ - 2y = - 60 
∴ y = 30 

∴ x = 80 - 30 
= 50
∴ There are 50 horses. 

Question: If x/y = 5/3 then (3x - 2y)/(3x + 2y) =?

Solution:
x : y = 5/3
⇒ x/y = 5/3
⇒ 3x/2y = (5 × 3)/(3 × 2) [Multiplying by 3/2]
⇒ 3x/2y = 15/6
⇒ (3x - 2y)/(3x + 2y) = (15 - 6)/(15 + 6)
⇒ (3x - 2y)/(3x + 2y) = 9/21 
⇒ (3x - 2y)/(3x + 2y) = 3/7
∴ (3x - 2y)/(3x + 2y) = 3 : 7

Question: A train passes a station platform in 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform?

Solution:
Speed = (54 × 1000)/3600 
= 15 m/sec

Length of the train = (15 × 20) m = 300 m
Let
the length of the platform be x meters.

Then,
(x + 300)/36 = 15
⇒ x + 300 = 540
∴ x = 240 meters

Question: Monir started a business investing Tk. 9000. After five months, Amir joined with a capital of Tk. 8000. If at the end of the year, they earn a profit of Tk. 6970, then what will be the share of Amir in the profit?

Solution:
Now as per the question,
Monir invested for 12 months
and Amir invested for (12 - 5) = 7 months.

So
Monir : Amir = (9000 × 12) : (8000 × 7)
= 108000 : 56000
= 108 : 56
= 27 : 14

Amir Ratio in profit will be = 6970 × (14/41)
= Tk. 2380

Question: A man sells two books at Tk. 120 each and by doing so he gains 25% on one book and loses 25% on the other. His loss on the whole in Tk. is = ?

Solution:
Sells price of two books = (120 × 2) = Tk. 240
Cost price of first book = (100/125) × 120
= Tk. 96

Cost price of second book = (100/75) × 120
= Tk. 160

∴ Loss = (160 + 96) - 240
= 256 - 240
= Tk. 16

Question: The radius of a circle is same as the diagonal of a square whose area is 16 sq. cm. The area of the circle is -

Solution:
Area of square = 16
Side of square = √16 = 4

Diagonal of square = 4√2

So, the radius of the circle is 4√2 cm

Area of circle = πr²
= π(4√2)²
= 32π cm²

Question: If xy(x + y) = 1 then = ?

Solution:
Given,
 xy(x + y) = 1
⇒ x + y = 1/xy

Now,
x³ + y³ - 1/x³y³
= x³ + y³ - (x + y)³
= x³ + y³ - {x³ + y³ + 3xy (x + y)}
= x³ + y³ - x³ - y³ - 3xy (x + y)
= - 3 . 1
= - 3

Question: Fahim and Robin invested in a business where the investment of Robin is double of Fahim. But Fahim immediately invested 15000 Tk. that brings him double the profit of Robin after one year. Robin's investment was -

Solution: 
Let initially the investment of Fahim is = X
So, the investment of Robin is = 2X

ATQ,
(X + 15000) : 2X = 2 : 1
⇒ 4X = X + 15000
⇒ 3X = 15000
∴ X = 5000 Tk.

Hence, the initial investment of Robin is = (2 × 5000) = Tk. 10000

Question: = ?

Solution:

Question: What percent of 700 is 2.1?

Solution:
Let,
the percent is x.

ATQ,
x% of 700 = 2.1
⇒ (x/100) × 700 = 2.1
⇒ 7x = 2.1
⇒ x = 2.1/7
∴ x = 0.3

Question: The difference between a number and its square is 72. What is the number?

Solution:
Let, the number be x.

ATQ,
x² - x = 72
⇒ x² - x - 72= 0
⇒ x² - 9x + 8x - 72 = 0
⇒ x(x - 9) + 8(x - 9) = 0
⇒ (x - 9) (x + 8) = 0
⇒ x - 9 = 0

Here,
x - 9 = 0
∴ x = 9

Question:


Solution:

Question: 10, 25, 45, 54, 60, 75, 80 
Which number doesn’t belong here?

Solution:
Each of the numbers except 54 is multiple of 5.

So, 54 doesn’t belong here.

Question: Calculate the H.C.F of 2x² - 8 and x² + 4x + 4

Solution: 
2x² - 8
= 2(x2 - 4)
= 2(x² - 2²)
= 2(x + 2)(x - 2)

And x² + 4x + 4
= (x² + 2 . x . 2 + 2²)
= (x + 2)²
= (x + 2)(x + 2)

∴ Required H.C.F = x + 2