Bank Math Master

Question: A is greater than B by 13; C is less than B by 7. If the ratio of A and C is 9 : 5, what is the value of B?

Solution:
Given,
A - B = 13
B - C = 7
and A : C = 9 : 5
⇒ A/C = 9/5
⇒ A = 9C/5

Now,
A - B + B - C = 13 + 7
⇒ A - C = 20
⇒ 9C/5 - C = 20
⇒ (9C - 5C)/5 = 20
⇒ 4C = 100
⇒ C = 25

So, B - 25 = 7
⇒ B = 32

Question: Tk 12000 are to be distributed between B and A such that B gets Tk 4000 less than A. The ratio of the amount received by A to that received by B is -

Solution:
A + B = 12000..........(i)
A - B = 4000...............(ii)

Now,
A + B + A - B = 12000 + 4000
⇒ 2A = 16000
⇒ A = 8000

So, 
B gets = 12000 - 8000 = 4000

∴ A : B = 8000 : 4000 = 2 : 1

Question: An iron rod that weights 30 kg is cut into two pieces so that one of these pieces weights 16 kg and 40 m long. If the weight of each piece is proportional to its length, how long is the other one is

Solution:
Total weight = 30 kg,
1st piece weight = 16 kg and Length = 40 m

So, 2nd piece weight = (30 - 16) = 14 kg
Let the Length of the 2nd piece = x m

Now,
16 : 14 = 40 : x
⇒ 16/14 = 40/x
⇒ x = (40 × 14)/16
⇒ x = 35

Question: The ratio of milk and water in a 48 liters mixture is 7 : 5. Find the quantity of milk to be added to the mixture in order to make this ratio 2 : 1.

Solution:
Here,
Milk = 48 × (7/12) = 28 L
Water = 48 × (5/12) = 20 L

Let x liters milk be added to the mixture.

Then,
(28 + x) : 20 = 2 : 1
⇒ (28 + x)/20 = 2/1
⇒ 28 + x = 40
⇒ x = 12

Question: The incomes of A and B are in the ratio 3 : 2 and their expenditure are in the ratio 5 : 3. If each saves Tk 1000, then, A's income can be-

Solution:
Let the income of A and B be 3x and 2x respectively.
Also, their expenditure is 5y and 3y.

According to the question,
3x - 5y = 1000 ------- (i)
2x - 3y = 1000 ---------- (ii)

Now, {(i) × 3} - {(ii) × 5}
⇒ 9x - 15y - 10x + 15y = 3000 - 5000
⇒ - x = -2000
⇒ x = 2000

Then, income of A = 3x = 3 × 2000 = Tk 6000

Question: Tk 1200 is lent out at 5% per annum simple interest for 3 years. Find the amount after 3 years.

Solution:
A = p(1 + nr)
= 1200{1 + (3 × 5)/100}
= 12000 × (115/100)
= 1380

Question: Interest obtained on a sum of Tk 6250 for 5 years is Tk 1250. Find the rate of interest.

Solution:
We know,
I = Pnr
⇒ r = I/pn
⇒ r = 1250/(5 × 6250)
⇒ r = (1250 × 100)/(5 × 6250)
⇒ r = 4%

Question: A sum of money at simple interest amounts to Tk 902 in 3 years and to Tk 962 in 5 years. The sum is- 

Solution:
S.I. for 2 year = Tk (962 - 902) = Tk 60
S.I. for 1 year = Tk 60/2 
S.I. for 3 years = Tk (60 × 3)/2= Tk 90

Therefore Principal = 902 - 90 = Tk 812

Question: A sum fetched a total simple interest of Tk 840 at the rate of 7 p.c.p.a. in 12 years. What is the sum?

Solution:
We know,
I = Pnr
⇒ P = I/nr
⇒ P = (840 × 100)/(7 × 12)
⇒ P = 1000

Question: A certain amount earns a simple interest of Tk 1350 after 15 years. Had the interest been 2% more, how much more interest would it have earned?

Solution:
We need to know the Simple interest, principal, and time to find the rate.
Since the principal is not given, So data is inadequate.

Question: What would be the compound interest accrued on an amount of Tk 10000 at the rate of 15% per annum in 2 years? 

Solution:
Compound Principal = 10000(1 + 15/100)²
= 10000 × (115/100)²
= (10000 × 115 × 115)/(100 × 100)
= 13225

Compound interest = 13225 - 10000 = 3225

Question: An amount of money in invested in a saving account for two years. It increases by Tk 420 in two years after annual compounding at the rate of 10% per year. What is the amount in Tk invested annually?

Solution:
Let, the principal be p

ATQ,
p + 420 = p(1 + 10/100)²
⇒ p + 420 = p(1 + 1/10)²
⇒ p + 420 = p(11/10)²
⇒ p + 420 = 121p/100
⇒ 121p = 100p + 42000
⇒ 21p = 42000
⇒ p = 2000

Question: The difference between compound interest and simple interest on a sum for 2 years at 8% is Tk 768. The sum is -

Solution:
Let the sum be p

Then, Compound Interest = p(1+ 8/100)² - p
= p(108/100)² - p
= p(27/25)² - p
= 729p/625 - p
= 104p/625

Again, Simple Interest = p × 2 × (8/100) = 4p/25

ATQ,
104p/625 - 4p/25 = 768
⇒ 4p/625 = 768
⇒ p = (768 × 625)/4
⇒ p = 120000