ব্যাংক ডেইলি কুইজ [লং কোর্সের অংশ]

Question: Tk. 1400 is lent out at 5% per annum simple interest for 4 years. Find the amount after 4 years.

Solution:
We know,
A = p(1 + nr)
= 1400 {1 + (4 × 5)/100}
= 1400 × (120/100)
= 1680

Question: The simple interest at m% for m years will be Tk. m on a sum of?

Solution:
Given, 
Rate of interest, r = m%
Interest, I = m Tk.
Time, n = m yeras.

We know,
I = Pnr
P = I/nr
= (m × 100)/(m × m)
= 100/m Tk.

Question: A sum of money amounts to Tk. 5200 in 5 years and to Tk. 5680 in 7 years at simple interest. The rate of interest per annum is?

Solution:
Simple interest for 2 years = (5680 - 5200) Tk.
= Tk. 480
∴ Simple Interest for 2 years =  Tk. 480
∴ Simple Interest for 5 years =  Tk. (480 × 5/2)
= Tk. 1200
∴ Principal = 5200 - 1200 = Tk. 4000.

We know, 
I = Pnr
⇒ r = I/Pn
 ⇒ r = (1200 × 100)/(4000 × 5)
∴ r = 6%

Question: If simple interest on a certain sum of money is Tk. 225 and the rate of interest per annum equals the number of years, then the rate of interest is-

Solution:
Let,
The rate of interest is x%
∴ Number of years is x

The interest of 1 year of 100 taka is x
∴ The interest of x years of 100 Taka is x × x = x²

ATQ,
x² = 225 
∴ x = 15

∴ The rate of interest is 15%

Question: At the rate of (17/2)% p.a. simple interest, a sum of Tk. 4800 will earn how much interest in 2 years 3 months?

Solution:
Time, n = 2 years 3 months = 2(1/4) years = 9/4 years.

We know,
I = Pnr
= 4800 × (9/4) × (17/2) × (1/100)
= Tk. 918

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 certain sum of money yields Tk. 1261 as compound interest for three years at 5% per annum. The sum is ?

Solution: 
Let the principal be Tk. P

Now,
1261 = P(1 + r)n - P
1261 = P(1 + 5/100)3 - P
1261 = P(105/100)3  - P
1261 = P(21/20)3 - P
1261 = P(9261/8000 - 1)
1261 = P (1261/8000)
1 = P/8000
P = 8000

Question: On a sum of money, the simple interest for 2 years is Tk. 660, while the compound interest is Tk. 696.30, the rate of interest being the same in both cases. The rate of interest is -

Solution:
Difference of Compound interest and Simple interest for 2 years = Tk. (696.30 - 660) = Tk. 36.30
Simple interest for 1 years = 660/2 = Tk. 330

Simple interest on 330 for 1 year = 36.30
Interst rate = {(36.30 × 100)/330}% = 11%

Question: Tahsan took a loan at simple interest rate of 6 p.c.p.a. in the first year and it increased by 1.5 p.c.p.a. every year. If he pays Tk. 8190 as interest at the end of 3 years, what was his loan amount?

Solution:
Let the loan amount be Tk. x

ATQ,
6% of x  + 7.5% of x + 9% of x = 8190
⇒ (6x/100) + (7.5x/100) + (9x/100) = 8190
⇒ 22.5x = 819000
∴ x = 36400

Question: In how many years will Tk. 2000 amounts to Tk. 2420 at 10% per annum compound interest?

Solution: 
Principal = Tk. 2000
Amount = Tk. 2420
Rate = 10% 

We Know
C = P(1 + r)ⁿ
2420 = 2000(1 + 10/100)ⁿ
2420/2000 = (11/10)ⁿ
121/100 = (11/10)ⁿ
(11/10)² = (11/10)ⁿ
n = 2

Question: Asif obtained an amount of Tk. 8376 as simple interest on a certain amount at 8 p.c.p.a. after 6 years. What is the amount invested by Asif?

Solution:
Simple interest, I = Tk. 8376
Rate of interest, r = 8%
Time, n = 6 years

We know,
I = Pnr
Or, P = I/nr
= (8376 × 100)/(6 × 8)
∴ P = 17450 Tk.