ব্যাংক ডেইলি কুইজ [লং কোর্সের অংশ]

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

Given,
Rate of interest, r = p%
Interest, I = p Tk.
Time, n = p years.

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

Question: Compound interest on a certain sum at the rate of 25% per annum after 2 years is Tk. 5,625. Find the simple interest on that sum at the rate of 12% per annum for 5 years.

Solution:

Let the sum be, P = x.
Rate of interest, r = 25%
Period, n = 2 years

We know,
Compound Principal, C = P(1 + r)ⁿ

∴ Compound interest = P(1 + r)ⁿ - P

Hence, compound interest = x(1 + 25/100)² - x = 5,625
⇒ x(5/4)² - x = 5,625
⇒ x(25/16) - x = 5,625
⇒ x(9/16) = 5,625
⇒ x = (5,625 × 16)/9
∴ x = 10,000 Tk.

Again,
Principal, P = 10,000 Tk.
Rate of interest, r = 12%
Period, n = 5 years

We know,
Simple Interest, I = Pnr/100

Hence, simple interest = 10,000 × 5 × 12/100
= 6,000 Tk.

Question: Find the difference between the simple interest and the compound interest at 10% per annum for 2 years on a principal of Tk 3,000.

Solution:
We know, Simple Interest, I = pnr
and Compound Principal, C = p(1 + r)ⁿ

Simple Interest = 3,000 × 2 × (10/100) = 600 Tk

Compound Principal = 3,000 × (1 + 10/100)²
= 3,000 × (110/100)²
= (3,000 × 110 × 110)/(100 × 100)
= 3,630

So, Compound interest = 3,630 - 3,000 = 630 Tk

So, difference = 630 - 600 = 30 Tk.

Question: The simple interest on a certain sum at 10% per annum for 2 years is Tk. 2,000. Find the compound interest on the same sum at the same rate for 2 years.

Solution:
We know,
I = Pnr

Given,
I = 2,000 Tk.
r = 10%
n = 2 years

∴ P = I/nr
= (2,000 × 100)/(2 × 10)
= 10000

Now,
Compound Principal = 10,000 × (1 + 10/100)²
= 10,000 × (110/100)²
= 10,000 × (11/10)²
= 10,000 × (121/100)
= 12,100

So, Compound interest = 12,100 - 10,000
= 2,100 Tk.

Question: If a sum triples in 15 years at simple interest, how much will it be in 10 years?

Solution:
Let, Principal = P

If the sum triples in 15 years,
Amount after 15 years = 3P

∴ Simple interest for 15 years = 3P - P
= 2P

So, Simple interest for 1 year = 2P/15

∴ Simple interest for 10 years = (2P/15) × 10
= 20P/15
= 4P/3

Amount after 10 years = P + 4P/3
= 7P/3

Hence, the sum will be = 7/3 times the principal.

Question: At what rate percent per annum will a sum of Tk. 64,000 amount to Tk. 85,184 in 3 years at compound interest?

Solution:
Given that,
Principal, P = Tk. 64,000
Amount, C = Tk. 85,184
Time, n = 3 years

We know,
C = P(1 + r)ⁿ
⇒ 85,184 = 64,000 × {1 + (r/100)}³
⇒ 85,184/64,000 = {1 + (r/100)}³
⇒ 1.331 = {1 + (r/100)}³
⇒ (1.1)³ = {1 + (r/100)}³

∴ 1 + (r/100) = 1.1
⇒ r/100 = 0.1
∴ r = 10%

Question: The difference between the compound interest and the simple interest on a sum at 10% per annum for 2 years is Tk. 800. Find the principal.

Solution:
Given That,
r = 10%

We know,
Difference between C.I and S.I for 2 years
= p(r/100)²
= P(10/100)²

ATQ,
P(10/100)² = 800
⇒ P(1/10)² = 800
⇒ P(1/100) = 800
⇒ P = 800 × 100
∴ P = 80,000

Hence, Principal = Tk. 80,000

Question: In how many years will Tk. 15,000 amount to Tk. 18,600 at 12% per annum simple interest?

Solution:
Principal, P = Tk. 15,000
Amount, A = Tk. 18,600

∴ Simple interest, I = A - P
= 18,600 - 15,000
= Tk. 3600

Rate of interest, r = 12%

We know,
I = Pnr
⇒ 3,600 = 15,000 × n × (12/100)
⇒ 3,600 = 15,000 × n × (3/25)
⇒ 3,600 = 1,800n
⇒ n = 3,600/1,800
∴ n = 2 years

Question: A sum of Tk. 24,000 is invested at 5% per annum simple interest. Find the interest earned in 4 years 6 months.

Solution:
Given that,
Principal, P = Tk. 24,000
Rate of interest, r = 5%
Time, n = 4 years 6 months
= 9/2 years

We know,
I = Pnr
⇒ I = 24,000 × 9/2 × 5/100
= 24,000 × 9 × 5/(2 × 100)
= 24,000 × 45/200
= 24,000 × 9/40
= 600 × 9
= Tk. 5,400

Question: What sum of money will amount to Tk. 2,500 in 4 years and to Tk. 2,900 in 6 years at simple interest?

Solution:
Increase in 2 years = 2,900 - 2,500
= Tk. 400

∴ Simple interest for 2 years = Tk. 400
∴ Simple interest for 4 years = Tk. (400 × 4/2)
= Tk. 800

Principal, P = 2,500 - 800
= Tk. 1,700

Question: A company increases salary of an officer at 20% per year. In 2025 an employee receives Tk. 43,200. What was his salary in 2023?

Solution:
Let, his salary in 2023 was 100 Tk.

At 20% increment,
Salary of 2024 is = 100(1 + 20/100)
= 120 Tk.

Salary of 2025 is = 120(1 + 20/100)
= 144 Tk.

Now,
144 Tk. corresponds to = 43,200 Tk.
1 Tk. corresponds to = 43,200/144 Tk.
∴ 100 Tk. corresponds to = (43,200 × 100)/144
= 30,000 Tk.

∴ His salary in 2023 was 30,000 Tk.

Question: A certain sum amounts to Tk. 12,100 in 2 years at 10% per annum compound interest. Find the principal.

Solution:

Principal, P = Tk. P
Amount, C = Tk. 12,100
Rate, r = 10%
Time, n = 2 years

We know,
C = P(1 + r)ⁿ
⇒ 12,100 = P × (1 + 10/100)²
⇒ 12,100 = P × (110/100)²
⇒ 12,100 = P × (11/10)²
⇒ 12,100 = P × (121/100)

∴ P = (12,100 × 100)/121
= 10,000

Hence, Principal = Tk. 10,000

Question: Rahim invested 20,000 Tk. in two 4 years term. He received 6% for the first term and 9% for the second term. What is the average rate of interest did he receive?

Solution:
In first term, he received = 20,000 × 4 × 6%
= 4,800 Tk.
In second term, he received = 20,000 × 4 × 9%
= 7,200 Tk.

So, total interest after 8 years, I = 4,800 + 7,200 = 12,000
P = 20,000
n = 8
r = ?

We know,
I = Pnr/100
∴ r = (100 × I)/Pn
= (100 × 12,000)/(20,000 × 8)
= 7.5%

∴ The average rate of interest Rahim received = 7.5%

Question: In how many years will Tk. 20,000 amount to Tk. 28,800 at 20% per annum compound interest?

Solution:
Given that,
Principal, P = Tk. 20,000
Amount, C = Tk. 28,800
Rate, r = 20%

We know,
C = P(1 + r)ⁿ
⇒ 28,800 = 20,000 × {(1 + (20/100)}ⁿ
⇒ 28,800 = 20,000 × (6/5)ⁿ
⇒ 28,800/20,000 = (6/5)ⁿ
⇒ 36/25 = (6/5)ⁿ
⇒ (6/5)² = (6/5)ⁿ

∴ n = 2 years

Question: A sum of Tk. 36,000 is invested at 8% per annum simple interest. Find the amount after 5 years.

Solution:
Principal, P = Tk. 36,000
Rate of interest, r = 8%
Time, n = 5 years

We know,
I = Pnr
⇒ I = 36,000 × 5 × 8/100
= 36,000 × 40/100
= 36,000 × 2/5
= Tk. 14,400

Amount, A = P + I
= 36,000 + 14,400
= Tk. 50,400