ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Question: Find the rate of interest if the amount after 2 years of simple interest on a capital of Tk. 1200 is Tk. 1440.

Solution:
Given,
Amount, A = Tk.1440
Principal, P = Tk.1200
Interest, I = Tk.(1440 - 1200) = Tk. 240
Time, n = 2 years.

We know,
I = Pnr
⇒ r = I/Pn
⇒ r = (240 × 100)/(1200 × 2) 
∴ r = 10%

Question: Hasan lent Tk. 400 to Asif for 2 years and Tk. 100 to Ratul for 4 years and received together from both Tk. 60 as interest. What is the rate of interest if the simple interest rate of both is the same?

Solution:
Let
rate is r%
The interest of Asif, I₁ = 400 × 2 × r%
The interest of Ratul, I₂ = 100 × 4 × r%

According to the question,
{(400 × 2 × r)/100} + {(100 × 4 × r)/100} = 60
Or, 8r + 4r = 60
Or, 12r = 60
Or, r = 60/12
∴ r = 5%

Question: Tk. 9800 are invested partly in 9% stock at 75 and 10% stock at 80 to have an equal amount of income. The investment in 9% stock is:

Solution:
Let,
the investment in 9% stock be tk. x.
Then, investment in 10% stock = tk.(9800 - x)

According to the question,
(9/75) × x = (10/80) × (9800 - x)
⇒ 3x/25 = (9800 - x)/8
⇒ 24x = (9800 × 25) - 25x
⇒ 24x + 25x = 245000
⇒ 49x = 245000
⇒ x = 245000/49
∴ x = 5000

Question: Which amount to be received after 3 years at the rate of 7% p.a. of simple interest on a sum of Tk. 2800?

Soution:
Given,
Rate of interest, r = 7%
Principal, P = 2800 Tk.
Time, n = 3 years.

∴ Total interest, I = Pnr
= (2800 × 3 × 7)/100
= 588 Tk.

So, amount = Principal + Simple interest
= (2800 + 588) Tk.
= 3388 Tk.

Question: A sum of money amounts to Tk. 735 in 3 years and Tk. 815 in 4 year. What's the sum?

Solution: 
Simple interest for 1 years = Tk. (815 - 735)
= Tk. 80
∴ Simple interest for 3 years = Tk.(80 × 3)
= Tk. 240

∴ Sum = (735 - 240)
= Tk. 495

Question: A man invested Tk. 14400 in Tk. 100 shares of a company at 20% premium. If the company declares 5% dividend at the end of the year, then how much does he get?

Solution:
Number of shares = 14400/120
= 120
Face value = Tk. (100 × 120)
= Tk. 12000

∴ Annual income = Tk. (5/100) × 12000
  = Tk. 600

Question: An amount of Tk. 6000 lent 5% per annum compounded interest for 2 years will become?

Solution:
Here,
Principal, P = Tk. 6000
Rate of interest, r = 5% = 5/100 = 1/20
Time, n = 2 years.

We know, 
C = P (1 + r)ⁿ
= 6000 × {1+ (1/20)}²
= 6000 × {(21/20) × (21/20)}
= 6000 × (441/400)
= 6615

∴ Compound Interest Tk. 6615.

Question: How many shares of market value Tk. 25 each can be purchased for Tk. 20,400, brokerage being 2% ? 

Solution:
CP of each share = Tk. (25 + 2% of 25) = Tk. 25.50. 

Number of shares= 20400/25.50
                             = 800

Question: A man buys tk. 20 shares paying 9% dividend. The man wants to have an interest of 12% on his money. The market value of each share is:

Solution: 
Dividend on Tk.20 = Tk. {(9/100) × 20}
= Tk. 9/5

Tk.12 is an income on Tk.100
∴ Tk. 9/5 is an income on = Tk.(100/12) × (9/5)
= Tk.15

Question:  How much time will it take for an amount of Tk. 450 to yield Tk. 99 as interest at 5.5% per annum of simple interest?  

Solution:
Here 
Principal = Tk. 450
S.I = Tk. 99 
R = 5.5% 

Time = (100 × 99)/(450 × 5.5)
= 4 years

Question: The population of a village decreases at the rate of 25% per annum. If its population 2 years ago was 24000, the present population is =?

Solution: 
Given that,
The population of a village 2 years ago = 24000
Rate of growth = 25% p.a.
Time period = 2 years

Present population = P{(1 - (r/100)}ⁿ
= 24000 × {1 - (25/100)}²
= 24000 × {1 - (1/4)}²
= 24000 × (3/4)²
= 24000 × (9/16)
= 13500

Question: A man invested Tk. 5940 in Tk.10 shares quoted at Tk. 8.25. If the rate of dividend be 12%, his annual income is: 

Solution:
Number of shares =(5940/8.25​) = 720

Face value = Tk.(720× 10)= Tk. 7200

Annual income = {(12/100) ​× 7200} = 864

Question: A certain sum of money becomes three times of itself in 20 years at simple interest. In how many years does it become double of itself at the same rate of simple interest?

Solution: 
Let,
the principal be P.
After 20 years, Amount = 3P
Simple interest, I = 3P - P = 2P

We know that,
SI = PRT/100
⇒ 2P = (P × R × 20)/100
⇒ R = 10%

Now,
Amount = 2P
SI = Amount - Principal
= 2P - P
= P

∴ SI = PRT/100
⇒ P = (P × 10 × T)/100
∴ T = 10 years

Question: A man took loan from a bank at the rate of 12% p.a. simple interest. After 3 years he had to pay. Tk. 5,400 interest only for the period, The principal amount borrowed by him was:

Solution:
Given,
Rate of interest, r = 12%= 12/100
Time, n = 3 years
Interest, I = Tk. 5400

Principal, P =? 

We know,
I = Pnr 
Or, 5400 = P × 3 × (12/100)
Or, P = (5400 × 100)/(3 × 12)
∴ P = 15000

Question: The simple interest on Tk. 36000 for the period from 5^th January to 31^st May 2013 at 9.5% per annum is =?

Solution:
Number of days = 27 + 28 + 31 + 30 + 31
= 147 days

We know,
I = Pnr
Or, I = (36000 × 147 × 9.5)/(100 × 365)
∴ I = Tk. 1,377.36

Question: A sum of Tk. 8000 will amount to Tk. 8820 in 2 years if the interest is calculated every year. The rate of compound interest is?

Solution:
Principal = Tk. 8000
Amount = Tk. 8820
Let Rate = R
Time = 2 years

By using formula, 
8000{1 + (R/100)}² = 8820
⇒ {1 + (R/100)}² = 8820/8000
⇒ {1 + (R/100)}² = 441/200
⇒ {1 + (R/100)}² = (21/20)²
⇒ 1 + (R/100) = 21/20
⇒ R/100 = (21/20) - 1
⇒ R/100 = 1/20
∴ R = 5%

Question: A certain sum of money is invested at an interest rate of 5% per annum and a second sum twice as large as the first is invested at 5.5% p. a. The total amount of interest earned from two investments together is Tk. 1000 per year and the interest is withdrawn every year. The first sum invested is: 

Solution:
Let 
Two sums Tk. x and Tk. 2x

Now
{(x × 5 × 1)/100} + {(2x × 5.5× 1)/100} = 1000
5x + 11x = 100000
16x = 100000
x = 6250

The first sum = Tk. 6250

Question: A man invests some money partly in 9% stock at 96 and partly in 12% stock at 120. To obtain equal dividends from both, he must invest the money in the ratio-

Solution:
For an income of Tk. 1 in 9% stock at 96,
investment = Tk. (96/9​) = Tk. 32/3​

For an income of Tk. 1 in 12% stock at 120,
investment = Tk. 120/12 = Tk. 10

∴ Ratio of investments =32/3 ​: 10 = 32 : 30 = 16 : 15

Question: The compound interest on a certain sum for 2 years at 10% per annum is Tk. 525. The simple interest on the same sum for double the time at half the rate percent per annum is:

Solution:
Let the sum be Tk. P.

ATQ,
P {1 + (10/100)}² - p = 525
⇒ P[{1 + (1/10)}² - 1] = 525
⇒ P{(11/10)² - 1} = 525
⇒ P{(121/100) - 1} = 525
⇒ P(21/100) = 525
⇒ P = (525 × 100)/21
∴ P = 2500

We know,
Simple interest, I = Pnr
= (2500 × 4 × 5)/100
= Tk. 500

Question: Mr. Anis took a loan of Tk.1400 with simple interest for as many years as the rate of interest. If he paid Tk. 686 as interest at the end of the loan period, what was the rate of interest?

Solution:
Given that n = r

We know,
S.I = (P × n × r)/100
⇒ 686 = (1400 × r × r)/100
⇒ 686 = 14r²
⇒ r² = 686/14
⇒ r² = 49
∴ r = 7% 

Question: By investing Tk. 1500 in a 10% stock , a man obtains an income of Tk. 150. Find the market price of the stock. 

Solution:
To earn Tk. 150 investment = Tk. 1500
To earn Tk. 10 investment = Tk. (1500 × 10)/150
= Tk. 100

Question: The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Tk. 1. The sum is:

Solution:
Let the sum be Tk. x.

Compound interest = x{1 + (4/100)}² - x
= x{1 + (1/25)}² - x
= x(26/25)² - x
= (676x/625) - x
= (676x - 625x)/625
= 51x/625

Simple interest = (x × 4 × 2)/100
= 2x/25

ATQ,
(51x/625) - (2x/25) = 1
⇒ (51x - 50x)/625 = 1
∴ x = 625

Question:  A 10% stock yields 8%. The market value of the stock is- 

Solution:
For an income of Tk. 8, investment = Tk. 100.
For an income of Tk. 10, investment = Tk. (100/8) × 10
                                                         = Tk. 125

Question: What will be the ratio of simple interest earned by certain amount at the same rate of interest for 9 years and that for 6 years? 

Solution:
Let 
The principal be P and  rate of interest be r% 

Required ratio = [(P × r × 9)/100]/[(P × r × 6)/100]
=9Pr/6Pr
= 9/6
= 3/2
= 3 : 2

Question: By investing in (50/3)% stock at Tk. 64, one earns Tk. 1500. The investment made is:

Solution:
To earn Tk. 50/3, investment = Tk. 64
To earn Tk. 1, investment = Tk. 64 × (3/50)
To earn Tk. 1500, investment = Tk. {64 × (3/50) × 1500}
= Tk. 5760

Question: A sum of Tk. 1550 was lent partly at 5% and partly at 8% p.a. simple interest. The total interest received after 3 years was Tk. 300. The ratio of the money lent at 5% to that lent at 8% is:

Solution:
Let
the sum lent at 5% be Tk. x and that lent 8% be Tk. (1550 - x).

ATQ,
(x × 5 × 3)/100 + {(1550 - x) × 8 × 3}/100 = 300
⇒ 15x + 37200 - 24x = 30000
⇒ - 9x = 30000 - 37200
⇒ - 9x = - 7200
⇒ x = 7200/9
∴ x = 800

∴ Required ratio = 800 : 750
= 16 : 15
 

Question: Tk. 6200 amounts to Tk. 9176 in 4 years at simple interest. If the interest rate is increased by 2% it would amount to how much? 

Solution:
Principle = Tk. 6200
Amount = Tk. 9176
Time = 4 years
Rate = ?
Interest = Tk. 9176 - Tk. 6200 = Tk. 2976

We know,
I = Pnr
Or, r = I/Pn
Or, r = (2976 x 100)/(6200 x 4)
= 297600/24800
= 12%

Now, if the rate is increased by 2% then final rate = 14%

Then, Interest = (6200 × 14 × 4)/100 = Tk. 3472


New amount =Tk.(6200 + 3472) = TK.9672