Bank Math Master

Question: If A and B are in the ratio 5 : 7 and B and C are in the ratio 14 : 15 then what is the ratio of A to C?

Solution: 
Given that, 
A : B = 5 : 7 and B : C = 14 : 15 

Now, 
(A/B) × (B/C) = (5/7) × (14/15)
⇒ A/C = (2/3)
∴ A : C = 2 : 3

Question: Kabir paid Tk. 9,600 as interest on a loan he took 5 years ago at the rate of 16% simple interest per annum. What was the principal amount (the original loan amount) he borrowed?

Solution: 
SI = Interest paid = Tk. 9600
R = Rate of interest = 16% per annum
T = Time = 5 years
P = Principal (the amount we need to find)

We know, 
SI = (P × R × T)/100
⇒ 9600 = (P × 16 × 5)/100
⇒ 9600 = (P × 80)/100
⇒ P = 96000/8
∴ P = 12000

∴ Kabir took a loan of Tk. 12000.

Question: If two times A is equal to three times of B and also equal to four times of C, then A : B : C is -

Solution:
2A = 3B
Or, B = 2A/3
and 2A = 4C
Or, C = A/2

Hence, A : B : C = A : 2A/3 : A/2
= 1 : 2/3 : 1/2
= 6 : 4 : 3

Question: Kamal invested Tk. 500 in EBL for 2 years and Tk. 300 in SIBL for 4 years. At the end of the period, he received a total of Tk. 220 as simple interest from both banks. What was the rate of interest per annum?

Solution:
Let the common rate of interest be r % per annum.

We know,
SI = (P × r × t)/100

Interest from EBL,
(500 × r × 2)/100 = (1000r)/100 = 10r

And, 
Interest from SIBL,
(300 × r × 4)/100 = (1200r)/100 = 12r

∴ Total interest received = Interest from EBL + Interest from SIBL
⇒ 220 = 10r + 12r 
⇒ 220 = 22r
⇒ r = 220/22
∴ r = 10

So the common rate of interest is 10% per annum.

Question: Ratio of two number A and B is 3 : 2. If we decrease 60 from A and Add 60 with B then ratio of A and B is 18 : 17. Find original value of A?  

solution:
Given that,
Ratio of A : B = 3 : 2
If A decreased by 60 and B increased by 60
And new ratio = 18 : 17

Let A = 3x and B = 2x

ATQ,
{3x - 60}/{2x + 60} = {18}/{17}
⇒ 17(3x - 60) = 18(2x + 60)
⇒ 51x - 1020 = 36x + 1080
⇒ 51x - 36x = 1080 + 1020
⇒ 15x = 2100
∴ x = 140

∴ Original value of A = 3x = 3 × 140 = 420

∴ The original value of A = 420.

Question: A person invests Tk. 3100 at a rate of 4% per annum under simple interest. After how many years will the total interest earned be Tk. 372?

Solution:
Given that,
Principal, P = Tk. 3100
Rate, r = 4%
Simple Interest, SI = Tk. 372

We know,
SI = (P × R × T) / 100
⇒ 372 = (3100 × 4 × n)/100
⇒ 372 = 124 × n
⇒ n = 372 ÷ 124
∴ n = 3 years

∴ Number of years = 3

Question: The monthly salaries of A, B and C are in the ratio 1 : 2 : 3. If C’s monthly salary is Tk. 1200 more than that of A, then what is B’s annual salary?

Solution:
Let the monthly salary of A, B and C be x, 2x and 3x
If C’s monthly salary is Tk. 1200 more than that of A, then

ATQ,
3x = x + 1200
⇒ 2x = 1200
⇒ x = 1200/2 = 600
∴ x = 600

Then, B’s monthly salary = 2x = 2 × 600 = 1200

∴ B’s annual salary = 1200 × 12 = Tk. 14400

Question: The compound interest on Tk. 2600 for 18 months at 10% per annum is-

Solution:
Interest after 1 year or 12 month = 2600 × 1 × (10/100)
= 260
New principal = 2600 + 260
= 2860

6 months = 1/2 year

Now, 
I = Pnr
= 2860 × (1/2) × (10/100)
= 143

∴ Total interest = 260 + 143 = Tk. 403

Question: Two numbers are in the ratio 5 : 4 and their difference is 10. Find the largest number.

Solution:
Let the two numbers be,
5x and 4x

According to the problem,
5x - 4x = 10
⇒ x = 10

So, the numbers are, 
5x = 5 × 10 = 50 and 4x = 4 × 10 = 40

∴ The largest number = 50

Question: In 4 years the simple interest on certain sum of money is 9/25 of the principal. The annual rate of interest is-

Solution:
Given that,
In 4 years, Simple Interest (SI) = 9/25​ of Principal (P)

We know,
SI = (P × r × n)/100
⇒ 9P/25 = (P × r × 4)/100
⇒ 9/25 = r/25
∴ r = 9

∴ Annual rate of interest = 9%

Question: How much should be added to each term of 4 : 7 so that it becomes 2 : 3?

Solution:
Given that,
Ratio of two numbers is 4 : 7
Let the number added to denominator and numerator be 'x' 

Now according to the question,
(4 + x) : (7 + x) = 2 : 3
⇒ (4 + x)/(7 + x) = 2/3
⇒ 12 + 3x = 14 + 2x
∴ x = 2 

∴ 2 will be added to make the term in the ratio of 2 : 3.

Question: What is the difference between the simple interest and the compound interest on a sum of Tk. 8000 for 2 years at the rate of 10% per annum when the interest is compounded yearly?

Solution:
Given that, 
Principal, P = Tk. 8000
Rate, r = 10% per annum
Time, n = 2 years

Wee know,
SI = (P × r × n)/100  
= (8000 × 10 × 2)/100  
= 160000/100  
= Tk. 1600

And,
Compound Interest (CI) – compounded annually
= P(1 + r)ⁿ - P
= P × (1 + 10/100)² - 8000
= 8000 × (1 + 1/10)²  - 8000
= 8000 × (1.1)²  - 8000
= 8000 × 1.21  
= 9680 - 8000
= Tk. 1680

Difference between CI and SI
= 1680 - 1600
= Tk. 80

Question: The ratio of the ages of A and B at present is 3 ∶ 1. Four years earlier the ratio was 4 ∶ 1. The present age of B is-

Solution:
Given that,
Ratio of present ages of A and B = 3 ∶ 1.
Ratio of their ages 4 years ago = 4 ∶ 1.

Let the present ages of A and B be 3x and x, respectively.
Four years ago,
Age of A = 3x - 4
Age of B = x - 4

AQT,
(3x - 4)/(x - 4) = 4/1
⇒ 3x - 4 = 4(x - 4)
⇒ 3x - 4 = 4x - 16
⇒ 3x - 4x = - 16 + 4
⇒ - x = - 12
∴ x = 12

∴ Present age of B = 12 years.

Question: Compound interest on a certain sum for 2 years at 10% per annum is Tk. 630. What would be the simple interest at the same rate and for the same time?

Solution:
Let, Principle = P
Compound Principle, C = P(1 + r)n
= P(1 + 10/100)2

ATQ,
C - P = 630
⇒ P(1 + 10/100)² - P = 630
⇒ P{(11/10)² - 1} = 630
⇒ P{(121/100) - 1} = 630
⇒ P(121- 100)/100 = 630
⇒ P(21/100) = 630
⇒ P = (630 × 100)/21
∴ P = 3000

Again,
I = Pnr
= 3000 × 2 × (10/100)
= 600

∴ The simple interest would be Tk. 600. 

Question: If a : b : c = 2 : 3 : 4 and 2a - 3b + 4c = 33, then the value of c is-

Solution:
Let,
 a = 2x, b = 3x and c = 4x  ; (since a : b : c = 2 : 3 : 4)

Now substitute into the equation,
2a - 3b + 4c = 33
⇒ 2(2x) - 3(3x) + 4(4x) = 33 
⇒ 4x - 9x + 16x = 33 
⇒ 11x = 33
∴ x = 3

Since c = 4x = 4 × 3 = 12

Question: A sum of money is borrowed and paid back in two annual instalments of Tk. 882 each allowing 5% compound interest. The sum borrowed was -

Solution:
Principle,
= (P.W. of Tk. 882 due 1 year hence) + (P.W of Tk. 882 due 2 years hence)
= [{882/(1 + 5/100)} + {882/(1 + 5/100)²}]
= [{(882 × 20)/21} + {(882 × 400)/441}]
= 840 + 800
= Tk. 1640

The sum borrowed was Tk. 1640.

Question: 94 is divided into two parts such that the fifth part of the first and the eighth part of the second are in the ratio 3 : 4. Find the first part.

Solution:
Let the two parts be x and 94 - x.

According to the problem,
(x/5) : (94 - x)/8 = 3 : 4
⇒ (x/5)/{(94 - x)/8} = 3/4
⇒ 8x/5(94 - x) = 3/4
⇒ 32x = 15(94 - x)
⇒ 32x = 15 × 94 - 15x
⇒ 47x = 15 × 94
⇒ x = (15 × 94)/47
∴ x = 30

∴ First part is 30

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: An iron rod that weights 24 kg is cut into pieces so that one of these pieces weighs 16 kg and is 34m long. If the weight of each piece is proportional to its length, how long is the other piece?

Solution:
Given that,
Total weight of rod = 24 kg
Cut into two pieces, one weighs 16 kg and is 34 m long
Weight ∝ Length
Let the length of the other piece be L₂​.
Its weight is W₂ = 24 - 16 = 8 kg.

Since weight is proportional to length,

W₁/L₁ = W₂/L₂
​⇒ 16/34 = 8/L₂
⇒ 16 × L₂ = 8 × 34
⇒ L₂ = (8 × 34)/16
∴ L₂ = 17 m

So the other piece is 17 meters long.

Question: A sum of money at simple interest doubles in 10 years. In how many years, at the same rate, will it be tripled?

Solution:
Given that,
Principal = P
Simple interest doubles the money in 10 years
 ∴ SI = P in 10 years.

We know,
SI = Prn/100
⇒ P = Prn/100
⇒ r × 10 = 100
⇒ r = 100/10
∴ r = 10% per annum 

Again,
To triple, total amount = 3P,  SI = 2P
2P = (P × 10 × n)/100
⇒ 2 = n/10
∴ n = 20

∴ Time taken to triple the amount is 20 years.