Bank Math Master

Question: Find the simple interest on BDT 15,000 at 8% per annum for 5 months.

Solution:
Principal, P = 15,000 Taka
Time, n = 5 months = 5/12 years
Rate of interest, r = 8% = 8/100

Simple Interest, I = P × n × r
= 15,000 × (5/12) × (8/100)
= (15,000 × 5 × 8)/(12 × 100)
= 600,000/1200
= 500

∴ The simple interest is Tk. 500.

Question: What is the compound amount of Tk. 3200 for 2 years at a rate of interest 5% per annum?

Solution:
Given,
Principal, P = 3200
Rate, r = 5% = 5/100 = 1/20
Time, n = 2 years

We know,
A = P(1 + r)ⁿ
  = 3200 × (1 + 1/20)²
  = 3200 × (21/20)²
  = (3200 × 21 × 21) / (20 × 20)
  = (3200 × 441) / 400
  = 1411200 / 400
  = 3528

∴ The compound amount is Tk. 3528.

Question: A certain principal amount, invested at simple interest, grows to Tk. 815 after 3 years and Tk. 854 after 4 years. What is the original principal amount?

Solution:
Given,
Amount after 3 years = Tk. 815
Amount after 4 years = Tk. 854
∴ Interest for 1 year = 854 - 815 = Tk. 39
∴ Interest for 3 years = 39 × 3 = Tk. 117
∴ Principal = 815 - 117 = Tk. 698

Question: The ratio of the present ages of a mother and son is 7 : 2. After 7 years, the ratio of their ages becomes 8 : 3. What will be the ratio of their ages after 11 years?

Solution:
Let,
The present age of the mother = 7x years
The present age of the son = 2x years

After 7 years, their ages will be:
Mother = (7x + 7) years
Son = (2x + 7) years

According to the question,
(7x + 7)/(2x + 7) = 8/3
⇒ 3(7x + 7) = 8(2x + 7)
⇒ 21x + 21 = 16x + 56
⇒ 21x - 16x = 56 - 21
⇒ 5x = 35
⇒ x = 35/5
∴ x = 7

Present age of mother = 7 × 7 = 49 years
Present age of son = 2 × 7 = 14 years

Now, after 11 years,
Mother's age = 49 + 11 = 60 years
Son's age = 14 + 11 = 25 years

∴ The ratio of their ages after 11 years,
= 60 : 25
= 12 : 5

Question: A sum of Tk. 30,000 yields a compound interest of Tk. 4347 when invested at 7% per annum. What is the investment period in years?

Solution:
Given,
Principal, P = 30000
Rate, r = 7% per annum
Compound Interest, CI = 4347

We know,
Amount, A = P + CI = 30000 + 4347 = 34347

Using the compound amount formula:
A = P(1 + r/100)ⁿ
⇒ 34347 = 30000 × (1 + 7/100)ⁿ
⇒ 34347 = 30000 × (107/100)ⁿ
⇒ (107/100)ⁿ = 34347/30000
⇒ (1.07)ⁿ = 1.1449
⇒ (1.07)ⁿ = (1.07)²
∴ n = 2 years

Question: What will be the total amount after 3 years if Tk. 1200 is invested at a simple interest rate of 5% annually?

Solution:
Here,
Principal, P = Tk. 1200
Rate of Interest, r = 5% = 5/100
Time, n = 3 years

We know,
Simple Interest, I = P × n × r
I = 1200 × 3 × (5/100)
I = 1200 × 3 × 5/100
I = 3600 × 5/100
I = 180 Tk.

Now, the Amount after 3 years, A = P + I
A = 1200 + 180
A = 1380 Tk.

∴ The amount after 3 years will be Tk. 1380.

Question: Two numbers are in the ratio 2 : 3. If 4 is subtracted from the first number, the ratio becomes 1 : 2. What are the numbers?

Solution:
Let the two numbers be: 2x and 3x

According to the question,
(2x - 4)/3x = 1/2
⇒ 2(2x - 4) = 3x
⇒ 4x - 8 = 3x
⇒ x = 8

∴ First number = 2 × 8 = 16
∴ Second number = 3 × 8 = 24

Question: The simple interest on a certain sum at 6% p.a. is Tk. 360 in 1 year. What will be the additional interest earned in 1 year if the rate increases to 7% p.a. on the same sum?

Solution:
Given,
Interest at 6% for 1 year:
Simple Interest, SI₁ = Tk. 360
Interest rate, R = 6%
Time, T = 1 year (initial)

We know,
SI₁ = PRT/100
⇒ 360 = (P × 6 × 1)/100
⇒ P = (360 × 100)/6
∴ P = Tk. 6000

So, the original sum (principal) is Tk. 6000.

Now,
Interest at 7% for 1 year:
SI₂ = PRT/100
⇒ SI₂ = (6000 × 7 × 1)/100
⇒ SI₂ = Tk. 420

Additional interest = SI₂ - SI_I
= 420 - 360
= Tk. 60

Question: A mixture contains copper, zinc, and tin in the ratio 2 : 3 : 5. If the total weight of the mixture is 50 grams, how many grams of tin are there in the mixture?

Solution:
Given,
Copper : Zinc : Tin = 2 : 3 : 5

Total parts = 2 + 3 + 5 = 10

Weight of tin = (5/10) × 50 = 25 grams

Question: A man bought some sugar and salt for Tk. 440. The ratio of the weights of sugar and salt is 5 : 2, and the price per unit weight of sugar and salt is in the ratio 6 : 7. What is the price of the total sugar?

Solution:
Given,
Weight ratio (Sugar : Salt) = 5 : 2
Price ratio per unit weight (Sugar : Salt) = 6 : 7

Total price ratio of sugar and salt = (Weight × Price per unit)
= 5 × 6 : 2 × 7
= 30 : 14
= 15 : 7

Total units = 15 + 7 = 22 units

Accordint to the question,
22 units = Tk. 440
⇒ 1 unit = 440/22
∴ 1 unit= Tk. 20

∴ Price of total sugar = 15 × 20 = Tk. 300

Question: In how many years will Tk. 750 amount to Tk. 900 at 4% simple interest per annum?

Solution:
Simple Interest = Amount - Principal
= 900 - 750
= 150

Here,
Principal, P = 750
Interest Rate, R = 4%
SI = 150
Time, T = ?

SI = PRT/100
⇒  T = (SI × 100)/(P × R)
= (150 × 100)/(750 × 4)
= 15000/3000
= 5 years

Question: If 2/3 of X = 75% of Y = 0.4 of Z, then find the ratio X : Y : Z.

Solution:
According to the question,
(2/3) × X = 75% of Y
⇒ 2X/3 = 75Y/100
⇒ 2X/3 = 3Y/4
⇒ X/Y = 9/8

Now,
75% of Y = 0.4 of Z
⇒(75Y/100) = (2Z/5)
⇒ (3Y/4) = (2Z/5)
⇒ 15Y = 8Z
⇒ Y/Z = 8/15

X : Y : Z = (9/8) : 1 : (15/8)
Multiply all terms by 8:
X = 9 × 1 = 9
Y = 1 × 8 = 8
Z = (15/8) × 8 = 15

Therefore,
X : Y : Z = 9 : 8 : 15

Question: Tk. 1,000 becomes Tk. 1,200 in 4 years at a certain rate of simple interest. If the rate of interest is increased by 2%, what amount will Tk. 1,000 become in 4 years?

Solution:
Principal, P = 1000
Interest, I = 1,200 - 1,000 = 200
Time, T = 4 years

SI = PRT/100
⇒ R = (SI × 100)/(P × n)
⇒ R = (200 × 100)/(1000 × 4)
⇒ R = 20000/4000
⇒ R = 5

∴ Interest rate, R = 5%

∴ New Interest rate = 5% + 2% = 7%

New interest at 7% for 4 years:
New interest = (1,000 × 7 × 4)/100
= (28,000)/100
= 280 Tk.

New amount = Principal + Interest
= 1,000 + 280
= Tk. 1,280

Question: The ratio of milk and water in a solution is 7 : 4. After adding 8 liters of water, the ratio of milk and water becomes 3 : 2. Find the final amount of water in the solution.

Solution:
Let the initial amount of milk = 7x liters
Let the initial amount of water = 4x liters

According to the question,
7x/(4x + 8) = 3/2
⇒ 2 × 7x = 3 × (4x + 8)
⇒ 14x = 12x + 24
⇒ 14x - 12x = 24
⇒ 2x = 24
⇒ x = 12

∴ Final amount of water = 4x + 8
= 4 × 12 + 8
= 48 + 8 = 56 liters

Question: How many years will Tk. 1500 need at 4% simple interest to earn the same interest as Tk. 2000 earns in 3 years at 5% simple interest?

Solution:
Interest from Tk. 2000 in 3 years at 5%:
SI = (P × R × T)/100
= (2000 × 5 × 3)/100
= 300 Tk.

Now,
Tk. 1500 to Produce the Same Interest (Tk. 300) at 4%:
Principal, P = Tk. 1500
Interest, SI = Tk. 300
Interest Rate, R = 4%
​ Time, T = ? years

SI = PRT/100
⇒ T = (SI × 100)/(P × R)
⇒ T = (300 × 100)/(1500 × 4)
⇒ T = 30000/6000
∴ T = 5 years