ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

P/S.I.=10/3
Let Principal = 10
S.I. for 5 year = 3
S.I. for 1 year = 0.6
Rate = S.I./Principal×100
Rate = 0.6/10×100
=6%

S.I. for 1 year=Tk.(854−815)=Tk.39
S.I. for 3 year=Tk.(39×3)=Tk. 117
∴Principal=Tk.(854−117)=Tk. 698

S.I.=Tk.(15500−12500)=Tk.3000
Rate=(100×3000)/(12500×4)%= 6%

Given,
S.I = 3 Principal Amount
=> 3A = A x 16 x R/100
By solving, we get
=> R = 18.75%

Let the interest rate be r%
We know that,
S.I = PTR/100
=> (1540 x 5 x r)/100  +  (1800 x 4 x r)/100 = 1788
=> r = 178800/14900 = 12%

(1500 x R1 x 3)/100 
=> 4500 (R1-R2) = 1350 
=> (R1-R2)= 1350/4500 = 0.3 %

Let the sum be Tk. P.
Then,[p(1+10/100)²-p]=525
Sum =Tk.2500
S.I.= Tk.(2500×5×4)/100
= Tk. 500

Amount = Tk.(30000+4347) = Tk.34347
let the time be n year
Then,
30000(1+7/100)ⁿ = 34347
(107/100)ⁿ = 34347/30000 = 11449/10000 = (107/100)²
n = 2year

Let the sum be Tk.P.then
P(1+R/100)³=6690…(i) and
P(1+R/100)⁶=10035…(ii)
On dividing,we get (1+R/100)³=10025/6690=3/2.
Substituting this value in (i),we get:
P×(3/2)=6690 or
P=(6690× 2/3)=4460
Hence,the sum is Tk.4460.

Let sum=Tk.x
C.I. when compounded half yearly = [x(1+10/100)⁴−x]=4641/10000
C.I. when compounded annually =[x(20/100)²−x]=11/25
4641/10000x−11/25x=482
=> x=20000

Time = 2 year 4 months = 2(4/12) year = 2(1/3) year.
Amount = Tk'. [8000 X (1+(15/100))² X (1+((1/3)×15)/100)]
=Tk. [8000 ×(23/20) × (23/20) ×(21/20)]
= Tk. 11109.
:. C.I. = Tk. (11109 - 8000)
= Tk. 3109.

Let Rajeev's present age be  x year.
Rajeev's age after 15 year = (x + 15) year.
Rajeev's age 5 year back = (x - 5) year
Then ATQ, 
x + 15 = 5 (x - 5)
x + 15 = 5x - 25
=> x = 10
Hence, Rajeev's present age = x = 10 year.

Let the mother's present age be x year.
Then, the person's present age = 2/5 x year.
(2x/5 + 8) = 1/2 (x + 8)
2(2x + 40) = 5(x + 8) 
=> x = 40.

Let current ages of X and Y correspondingly, is 6A & 5A
Given: 6A + 5A = 44
=> A = 4
Proportion of ages after 0.8 decades will be
6A + 8 : 5A + 8
32:28 (or) 8:7

Let number of boys = x,
Let number of girls = y
Total numbeTk of students = x + y
(x + y) × 15.8 = 16.4x + 15.4y 
0.6x = 0.4y
x/y = 0.4/0.6 = 2/3
Ratio of boys and girls in the class is 2:3

Let A's age 10 year ago = x year. 
Then, B's age 10 year ago = 2x year.
(x + 10) / (2x+ 10) = 3/4 
=> x = 5.
So, the total of their present ages 
=(x + 10 + 2x + 10) 
= (3x + 20) 
= 35 year.

Let Siya's age = x
Then Riya's age = 4x
But given Riya's age = 24
=> 4x = 24 
=> x = 6
Hence Siya's age = 6 years
=> The age difference = 24 - 6 = 18 years.

Let C's age be x year. 
Then, B's age = 2x year. 
A's age = (2x + 2) year. 
(2x + 2) + 2x + x = 27 
5x = 25 
=> x = 5 
Hence, B's age = 2x = 10 year.

Let the age of sushma be x and 
the age of her son is y
Then five year before x-5=5(y-5) ...(1)
Five year hence x+5 = 3(y+5)-8 .....(2)
By soving (1) & (2), we get
5y - 15 = 3y + 7
y = 11 
=> x = 35
Therefore, the age of Sushma = 35 and her son = 11.

Let the age of the youngest child is 'X' year.
Then the child elder than the yongest is 'X + 3' years
Since each have 3years difference upto 5 children
And given that
=> X + X+3 + X+6 + X+9 + X+12 = 50
=> 5X + 30 = 50
=> X = 4 years.

Let father's age be x year and son's age be y year.
According to question,
2(x+y) = 8y _______(I)
and (x+y)/2 = 30
=> x+y = 60 year_______(II)
From equation (I) and (II)
8y = 120
y = 15 year,
Hence x = 45 year.

Sum of ages of husband, wife and child= (24 × 3) + 9 = 81 year
Sum of ages of wife and child 
=> 25 × 2 + 10 = 50 + 10 = 60 year
Age of husband = 81 - 60 = 21 year

Let the average of 8 men be x.
Age of new man be N.
Total age of 8 men = 8x
⇒ x + 4 = ((8x – 30) + N)/8
⇒ 8x + 32 = 8x – 30 + N
⇒ N = 32 + 30
⇒ N = 62

From the given data,
when my age is 24, my mothers age is double of my age 
=> 48 years is my mothers age
=> Difference is 48 - 24 = 24 year.
When my age is 44
=> My mother is 44 + 24 = 68 year aged.