ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Question: The simple interest on a certain sum at the rate of 5% per annum for 3 years and 7 years differs by Tk. 400. What is the principal amount?

Solution:
Let the principal amount be Tk. P
Then, ATQ
400 = {(P × 5 × 7)​/100} - {(P × 5 × 3​)/100}
⇒ 400 = (35P - 15P)/100
⇒ 40000 = 20P
⇒ P = 40000/20
∴ P = 2000

So the principal amount is Tk. 2000

Question: At the rate of 15% simple interest per annum, how much profit will be earned on Tk. 900 as principal in 7 years?

Solution:
Given,
Sum P = Tk. 900
Time n = 7 years
Rate r = 15%
= 15/100
= 3/20

We know,
Profit = Prn
= 900 × 3/20 × 7
= 945

Question: What is the difference between simple and compound interest at 8% per annum on a sum of Tk. 12,500 at the end of 2 years?

Solution:
Here,
Principal, P = 12500 Tk.
Interest Rate, R = 8%
Time, T = 2 years

We know,
Simple Interest, SI = PRT/100
= (12500 × 8 × 2)/100
= 200000/100
= 2000 Tk.

For Compound Interest,
Amount, A = P{1 + (R/100)}^T
= 12500{1 + (8/100)}²
= 12500(1.08)²
= 12500 × 1.1664
= 14580 Tk.

∴ CI = A - P
= 14580 - 12500
= 2080 Tk.

∴ Difference between compound interest and simple interest,
= 2080 - 2000
= 80 Tk.

Question: An accurate clock shows 3:00 PM. Through how many degrees will the hour hand rotate when the clock shows 9:00 PM?

Solution:
From 3.00 pm to 9.00 pm,
total time = 9 - 3 = 6 hours

We know,
Angle traced by hour hand in 12 hours =  360°

∴ Angle traced by hour hand in 6 hours, 
= (360 × 6)/12
= 180°

Question: What is the difference between the simple interest on a principal of Tk. 1000 being calculated at 7% per annum for 2 years and 6% per annum for 3 years?

Solution:
Given that,
Principal, P = Tk. 1000
Rate, r₁ = 7% and Time, n₁ = 2 years
Rate, r₂ = 6% and Time, n₁ = 3 years

We know,
I = Prn/100
1st case:
I₁ = Prn/100 = (1000 × 7 × 2)/100 = 140

2nd case:
I₂ = Prn/100 = (1000 × 6 × 3)/100 = 180

∴ Difference = 180 - 140 = Tk. 40

Question: It was Tuesday on January 1, 2008. What was the day of the week on Jan 1, 2009?

Solution: 
The year 2008 is a leap year. So, it has 2 odd days.
Given,1st day of the year 2008 is Tuesday 
So, 1st day of the year 2009 is 2 days beyond Tuesday.
Hence, it will be Thursday.

Question: Find the amount if Tk. 2000 is invested at 10% compound interest p.a. for 3 years.

Solution: 
Given,
P = Tk. 2000
r = 10%
n = 3 years

We know,
C = P(1 + r)ⁿ
= 2000{1 + (10/100)}³
= 2000(1 + 1/10)³
= 2000(11/10)3
= 2000 × 1.1 × 1.1× 1.1
= 2662

Question: A sum of money amounts to Tk. 15000 in 4 years at 25% simple interest per annum. Find the sum. 

Solution:
Given, 
A = 15000 
T = 4
R = 25%

We know,
SI = P × R × T
= P × (25/100) × 4
= P

A = P + SI 
⇒ A = P + P
⇒ A = 2P
⇒ 15000 = 2P
⇒ P = 15000/2
∴ P = 7500

Question: A sum of money at simple interest amounts to Tk. 815 in 3 years and to Tk. 854 in 4 years. The sum is-

Solution:
Simple interest for 1 year = Tk. (854 - 815)
= Tk. 39

∴ Simple interest for 3 years = Tk.(39 × 3)
= Tk. 117

∴ Sum = (815 - 117)
= Tk. 698

Question: In how many years Tk. 1000 will become Tk. 1331 at compound interest rate of 10% per annum?

Solution:
Given,
P = Tk. 1000
C = Tk. 1331
r = 10% p.a

We know,
P(1 + r)ⁿ = C
⇒ 1000(1 + 10%)ⁿ = 1331
⇒ (1 + 1/10)ⁿ = 1331/1000
⇒ (11/10)ⁿ = (11/10)³
∴ n = 3

∴ required time = 3 years

Question: A man buys Tk. 20 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. 20 = Tk. (9/100) × 20 = Tk. 9/5

Tk. 12 is an income on Tk. 100.
∴ Tk. 9/5 is an income on = Tk. (100 × 9)/(12 × 5)
= Tk.15

Question: A town experiences a 10% decrease in its population every year. If the population was 20,000 two years ago, what is the population now?
(এক শহরের জনসংখ্যা প্রতি বছরে ১০% কমে যায়। যদি দুই বছর আগে জনসংখ্যা ২০,০০০ ছিল, তাহলে বর্তমানে জনসংখ্যা কত হবে?)

Solution:
বর্তমান জনসংখ্যা = 20000(1 - 10/100)² {[C = p(1 - r)ⁿ] অনুসারে।}
= 20000 × 0.9 × 0.9
= 16,200

Question: Find out the wrong number in the following series:
5, 10, 17, 26, 36, 50, 65

Solution:
Given,
The following series: 5, 10, 17, 26, 36, 50, 65

The difference between consecutive numbers of the given series are respectively = 5, 7, 9, 11, 13 and 15.
Therefore, 26 + 11 = 37
But in this question it is given 36.

So 36 is wrong number.

Question: If the simple interest on Tk. M at M% per annum for 4 years is Tk. M, what is the value of M?

Solution:
Given,
P = M
r = M% = M/100
n = 4 years
I = M

We know,
I = Prn
⇒ M = M × (M/100) × 4
⇒ M = M × M/25
∴ M = 25

Question: What word can be formed by arranging the letters of 'AEEHLPTN'?

Solution:
By rearranging the letters 'AEEHLPNT', the name of an animal (ELEPHANT ⇒ Elephant) can be formed.

Question: Find the initial sum that grows to Tk. 1500 in 5 years and Tk. 1620 in 7 years at simple interest.

Solution:
in two years the interest becomes = 1620 - 1500 = 120Tk.
in 5 years, interest = (120/2)/5 =300 Tk.

so , the initial sum is = 1500 - 300 = 1200 Tk.

Question: How many times are the hands of a clock at right angle in a day?

Solution:
In 12 hours, they are at right angles 22 times.
∴ In 24 hours, they are at right angles 44 times.

Question: Some principal becomes Tk. 1,560 as profit-principal in 3 years and Tk. 1,920 as profit-principal in 6 years. Find the rate of profit.

Solution: 
Given,
Principal + Profit for 3 years = 1560
Principal + Profit for 6 years = 1920

So the profit of 3 years I = Tk. (1920 - 1560)
= Tk. 360

So the principal P = Tk. (1560 - 360)
= Tk. 1200

∴ The rate of profit r = (I × 100)/(P × n)
= (360 × 100)/(1200 × 3)
= 10

Question: If England is written as 1234526 and France is written as 785291, then Greece is written as-

Solution: 
England is written as 1234526
E = 1
n = 2
g = 3
l = 4
a = 5
n = 2
d = 6

France is written as 785291
F = 7
r = 8 
a = 5
n = 2
c = 9
e = 1

So, GREECE will be = 381191

Question: The difference between compound interest and simple interest on a certain sum for 2 years at 12% per annum is 36 Taka. What is the principal amount?

Solution: 
For 2 years, the difference between compound interest (CI) and simple interest (SI) is given by:

Difference = P × (r/100)t
where, 
P = Principal amount
r = interest rate = 12%
Time = t = 2 years
Difference = 36 Taka

So, we get, 
36 = P × (12/100)2
36 = P × (144/10000)
∴ P = 2500 Taka

Question: A ferry can carry 24 buses or 36 cars at a time. If there are 18 buses on the ferry, how many cars can be loaded onto it?

Solution:
Here,
24 buses = 36 cars
∴ 1 bus = 36/24 cars
∴ 18 buses = (36 × 18)/24 cars
= 27 cars

∴ Required number of cars = 36 - 27 = 9 cars

Question: A sum of money doubles itself in 10 years at simple interest. What is the annual interest rate?

Solution: 
Let,
Principal amount = P
Sum of amount = 2P
∴ Interest, I = 2P - P = P
Time, n = 10 years
Rate of interest = r

We know, 
I = Pnr
∴ r = I/Pn
= P/(P × 10)
= 1/10 × 100%
= 10% 

Question: If Tk. 3000 is loaned for 4 months at a 4.5% annual rate, how much interest is earned?

Solution:
P = 3000
n = 4 months = 4/12 years = 1/3 year
r = 4.5%

I = Pnr
= {3000 × (1/3) × (4.5)}/100
= 45