Bank Math Master

Question:  A mixture contains juice and sugar syrup in equal proportion. If a new mixture is created by adding this mixture and sugar syrup in the ratio 1 : 3, then the ratio of  juice and sugar syrup in the new mixture is - 

Solution: 
Let,
New mixture is = 12 liters
Old mixture = 12 × (1/4) = 3 liters 

Added sugar syrup = 12 - 9 = 9 liters 
New sugar syrup = 9 + (3/2) = 10.5 liters 

Ratio of  juice and sugar syrup = 1.5 : 10.5 
= 15 : 105 
= 1 : 7

Question: In what ratio must a mixture of 25% alcohol strength be mixed with that of 50% alcohol strength so as to get a mixture of 40% alcohol strength?

Solution: 
Let, X contains 25% alcohol strength and
Y contains 50% alcohol strength 

ATQ,
(25% of X) + (50% of Y) = 40% of (X + Y)
⇒ 25X + 50Y = 40X + 40Y
⇒ 15X = 10Y
∴ X : Y = 2 : 3

Question: Poltu divided one-third of his money in a 2 : 3 ratio between Ratul and Shovon, and the remainder in the same ratio between Nitul and Sanhar. If Sanhar received 360 Tk, what was Poltu's total amount?

Solution: 
If Sanhar received 360 Tk,
Nitul received = (360/3) × 2 Tk
= 240 Tk

Total amount Nitul and Sanhar received = 360 + 240 Tk
= 600 Tk [which is {1 - (1/3)} or 2/3 of total amount]

∴ Total amount = (600/2) × 3 Tk
= 900 Tk

Question: The ratio of milk and water in a solution is 25 : 12 and after adding 9 liters of water in it the ratio of milk and water becomes 5 : 3, then find the final amount of water in the final solution.

Solution: 
Let,
The initial amount of milk be 25x liters
and the amount of water 12x liters.

Ratio of milk and water after adding 5 liters 
25x/(12x + 9) = 5/3
⇒ 75x = 60x + 45
⇒ 15x = 45
∴ x = 3

∴ Final amount of water in solution = 12x + 9 liters.
= 36 + 9 liters.
= 45 liters.

Question: Two containers have mixtures of milk and water, respectively, in the ratios 5 ∶ 2 and 9 ∶ 5. In what ratio should the contents be mixed so that the ratio of milk to water in the final mixture is 2 ∶ 1? 

Solution :
Let,
P unit of the first mixture is added to Q unit of the second mixture.

So, in P unit of first mixture,
Amount of milk present = (5/7) × P = 5P/7
Amount of water present = (2/7) × P = 2P/7

In Q unit of second mixture,
Amount of milk present = (9/14) × Q = 9Q/14
Amount of water present = (5/14) × Q = 5Q/14

ATQ,
{(5P/7) + (9Q/14)}/{(2P/7) + (5Q/14)} = 2/1
⇒ {(10P + 9Q)/14}/{(4P + 5Q)/14} = 2
⇒ 10P + 9Q = 8P + 10Q
⇒ 2P = Q
∴ P : Q = 1 : 2 

Question: Find the difference between the simple interest and the compound interest at 5% per annum for 2 years on a principal of Tk 4000.

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

∴ Simple Interest = 4000 × 2 × (5/100) = 400 Tk

Compound Principal = 4000 × {1 + (5/100)}²
= 4000 × (105/100)²
= 4000 × (105/100) × (105/100)
= 4410

So, Compound interest = 4410 - 4000 = 410 Tk
∴ Difference = 410 - 400 = 10 Tk

Question: Find the compound interest on Tk 8000 at 15% interest per annum for 2 years, compounded annually-

Solution:
Compound Principal for 2 years = 8000{1 + (15/100)²
= 8000(115/100)²
= (8000 × 115 × 115)/(100 × 100)
= 10580 taka

Compound interest = 10580 - 8000 tk
= 2580 tk.

Question: A sum of money at simple interest amount to tk. 4400 in 2 years and to tk. 4700 in 5 years at a rate of:

Solution: 
Simple interest for 3 years = (4700 - 4400) Tk.
= Tk. 300
Now, Simple Interest for 3 years =  Tk. 300
∴ Simple Interest for 5 years =  Tk. 300 × (5/3)
= Tk. 500

∴ Principal = 4700 - 500 = Tk. 4200

We know, 
I = Pnr
⇒ r = I/Pn
 ⇒ r = (500 × 100)/(4200 × 5)
∴ r = 2.3%

Question: At what rate percent per annum will the simple interest on a sum of money 3/5 of the amount in 5 years?

Solution:
Let the principal be p Tk
So, interest = 3p/5 Tk
Time = 5 years
rate of interest, r =?

Now,
I = pnr
r = I/pn
= (3p/5)/(p × 5)
= (3/25) × 100%
= 12%

Question: The simple interest on a certain sum of money for 5 years at 10% per annum is Tk. 150 less than the simple interest on the same sum for 7 years at  8% per annum. Find the sum.

Solution: 
Let,
sum = x taka 

ATQ, 
{x × (8/100) × 7} - {x × (10/100) × 5} = 150 
⇒ (14x/25) - (x/2) = 150 
⇒ (28x - 25x)/50 = 150 
⇒ 3x/50 = 150 
⇒ x/50 = 50 
∴ x = 2500

Question: A stick is divided in the ratio 4 : 3 : 1. If the smallest part measures 7.5 cm, what is the length of the longest part?

Solution: 
Given, 
The stick is divided in the ratio 4 : 3 : 1. 

Let, 
Broken parts are 4x, 3x and x cm.

ATQ, 
x = 7.5
∴ 4x = 4 × 7.5
= 30 

Therefore, the largest part = 30 cm

Question: The ratio of present age Anik and Babul is 12 : 7 . After 3 years their ratio will be 30 : 20 then find the present age of Anik? 

Solution:
Let,
present age of Anik = 12x
and present age of Babul = 7x 

ATQ,
(12x + 3)/(7x + 3) = 30/20
⇒(12x + 3) × 20 = (7x + 3) × 30
⇒ 240x + 60 = 210x + 90 
⇒ 30x = 30
⇒ x = 1

∴ Present age of Anik = 12 × 1 years
= 12 years

Question: What is the ratio of 6 inches to 6 feet?

Solution: 
We know,
1 feet = 12 inches
So, 6 feet = 6 × 12
= 72 inches

Now, 
8 inches : 6 feet = 6 : 72 = 1 : 12

Question: When three-fifths of a number are subtracted from two-thirds of the number, the result is 18. What's the number?

Solution:
Let,
The number be n

ATQ,
(2/3) × n - (3/5) × n = 18
⇒ (2n/3) - (3n/5) = 18
⇒ (10n - 9n)/15 = 18
⇒ n/15 = 18
∴ n = 270

Question: A and B work together as business partners. A contributes 1/3 of the capital over 15 months, while B receives 2/5 of the profit. Approximately how long was B's money used?

Solution:
Let the total profit be Tk. z
Then,
B's share of profit = Tk. 2z/5, 
A's share of profit = Tk. z - (2z/5 ) = Tk. 3z/5
 
∴ A : B = (3z/5) : (2z/5)
= 3 : 2

Let the total capital be Tk. x and suppose B's money was used for n months.

ATQ,
{(x/3) × 15}/{(2x/3) × n}= 3/2
⇒ 5x × (3/2xn) = 3/2
⇒ 15/2n = 3/2
⇒ 6n = 30
∴ n = 5

∴ B's money was used for 5 months.

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Question: A man buys Tk. 50 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. 50 = Tk. (9/100) × 50
= Tk 9/2

Tk.12 is an income of Tk.100.
∴ Tk. 9/2 is an income on = Tk. (100/12) × (9/2)
= Tk. 37.5

Question: If interest is compounded every six months, how much interest will Tk. 4000 earn at a 10% annual rate in a year?

Solution:
Here,
Interest rate, r = 10% per year
= (10/2)% per 6 months
= 5% per 6 months
P = Tk. 4000
n = 2

∴ C = P(1 + r)ⁿ
= 4000 {1 + (5/100)}²
= 4000 × (105/100) × (105/100)
= 4000 × (21/20) × (21/20)
= 4410

∴ Interest is = Tk. 4410 - 4000
= Tk. 410

Question: At the rate of 12%  simple interest, a sum of Tk. 3000 will earn how much interest in 2 years 3 months?

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

We know,
I = Pnr
= 3000 × (9/4) × (12/100)
= 30 × 27
= Tk. 810

Question: A village's population is decreasing by 20% per year. If the population was 5000 two years ago, what is the current population?

Solution: 
Given that,
The population of a village 2 years ago = 5000
Rate of growth = 20% p.a.
Time period = 2 years

Present population = P{(1 - (r/100)}ⁿ
= 5000 × {1 - (20/100)}²
= 5000 × {1 - (1/5)}²
= 5000 × (4/5)²
= 5000 × (16/25)
= 3200

Question: An amount has increased by 80% over 16 years at simple interest. What is the compound interest on Tk. 12,000 after two years at the same rate?

Solution:
Let,
simple interest r% 

I = pnr/100
⇒ I/p = nr/100
⇒ 0.8 = 16r/100 
⇒ 8/10 = 16r/100 
∴ r = 5% 

The compound interest = 12000 (1+5/100)² - 12000 
= 12000 × 105/100 × 105/100 - 12000 
= 12000 × 21/20 × 21/20 - 12000 
= 12000 × 441/400 - 12000 
= 13230 - 12000 
= Tk. 1230