ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Question: There is 75% increase in an amount in 5 years at simple interest. What will be the compound interest of TK. 16,000 after 2 years at the same rate?

Solution:
Let,
P = Tk. 100
Then, S.I = Tk. 75
and n = 5 years

So the rate r = (100 × 75)/(100 × 5)
= 15%

Now, P = Tk. 16,000
n = 2 years
r = 15% p.a.
= 3/20 p.a

C.I = 16,000 × {(1 + 3/20)² - 1}
= 16,000 × {(23/20)² - 1}
= 16,000 × (529/400 - 1)
= 16000 × 129/400
= 5,160

Question: Rana took a loan of Tk. 7500 with simple interest for as many years as the rate of interest. If he paid Tk. 1875 as interest at the end of the loan period, what was the rate of interest?

Solution:
Let,
rate r = R%
and time n = R years

ATQ,
Prn = I
⇒ (7500 × R × R)/100 = 1875
⇒ 75R² = 1875
⇒ R² = 1875/75
⇒ R² = 25
∴ R = 5

Question: Which number replaces the question mark?

Solution:
Given,
1^st number = 1
2^nd number = 1 × 5 + 5 = 10
3^rd number = 10 × 5 + 5 = 55
4^th number = 55 × 5 + 5 = 280
∴ 5^th number = 280 × 5 + 5 = 1405

Question: Mr. Sohan is planning to buy a mobile after two years time from now. He expects that after two years he will need Tk. 18150 to buy the mobile. How much money does Mr. Sohan need to invest now at 10% interest rate compounded annually?

Solution:
Given,
Compounded Amount = Tk. 18150
Time n = 2 years
Rate r = 10% = 10/100 = 1/10

We know,
Compounded Amount = P(1 + r)ⁿ
⇒ 18150 = P(1 +1/10)²
⇒ 18150 = P(11/10)²
⇒ 18150 =121P/100
⇒ 121P =18150 × 100
⇒ P =1815000 ÷ 121
∴ P = 15000

Question: The compound interest on a certain sum for 2 years at 12% per annum is Tk. 795. The simple interest on the same sum for double the time at half the rate percent per annum is-

Solution: 
Let
The sum be Tk. P

ATQ,
P{1 + (12/100)}² - P = 795
⇒ P[{1 + (3/25)}² - 1] = 795
⇒ P{(28/25)² - 1} = 795
⇒  P{(784/625) - 1} = 795
⇒  P(159/625) = 795
⇒ P = (795 × 625)/159
∴ P = 3125

So The simple interest on the same sum for double the time at half the rate percent per annum is-
SI = 3125 × 4 × 6%
= 3125 × 4 × 6/100
= 750

Question: In how many years will Tk. 4000 amounts to Tk. 5324 at 10% per annum compound interest?

Solution:
Given,
P = Tk. 4000
Compound amount = Tk. 5324
Rate r = 10% = 1/10

We know,
Compound amount = P(1 + r)ⁿ
⇒ 5324 = 4000(1 + 1/10)ⁿ
⇒ 5324/4000 = (11/10)ⁿ
⇒ 1331/1000 = (11/10)ⁿ
⇒ (11/10)³ = (11/10)ⁿ
∴ n = 3

Question: A person borrows Rs. 5000 for 2 years at 4% p.a. simple interest. He immediately lends it to another person at 6% p.a compound interest for 2 years. Find his total gain in the transaction.

Solution:
Given,
P = Tk. 5000
n = 2 years
r = 4% = 4/100 = 1/25

So the simple interest = 5000 × (1/25) × 2
= Tk. 400

Again,
compound interest rate r = 6% = 6/100 = 3/50

So,
The compound interest = P(1 + r)ⁿ - P
= 5000(1 + 3/50)² - 5000
= {5000 × (53/50)²} - 5000
= (5000 × 2809/2500) - 5000
= 5618 - 5000
= 618

So the total gain in the transaction = (618 - 400)
= Tk. 218

Question: Find the minimum number of straight lines required to make the given figure.

Solution:

The figure may be labelled as shown.


The horizontal lines are IJ, AB, EF, MN, HG, DC and LK = 7 in number.
The vertical lines are AD, EH, IL, FG, BC and JK = 6 in number.
So, the minimum number of straight lines required to make the given figure = 7 + 6= 13

Question: A sum of money on compound interest amounts to Tk. 13200 in 3 years and Tk. 12000 in 2 years. The rate of interest per annum is?

Solution:
Let,
Sum = Tk. P
Rate = r% = r/100

ATQ,
P(1 + r/100)² = 12000............(1)
P(1 + r/100)³ = 13200.........(2)

Dividing equation (2) by (1):
1 + r/100 = 13200/12000
⇒ (100 + r)/100 = 11/10
⇒ 100 +r = (11 × 100)/10
⇒ 100 +r = 110
⇒ r = 110 - 100
∴ r = 10

Question: Due to increase the rate of simple interest from 5% to 7%, the income of Rahim was increased by Tk. 1200 in 3 years. How much was his principal?

Solution:
Let,
His principal = P

ATQ,
{P × (7/100) × 3} - {P × (5/100) × 3} = 1200
⇒ (21P/100) - (15P/100) = 1200
⇒ (21P - 15P)/100 = 1200
⇒ 6P/100 = 1200
⇒ 6P = 1200 × 100
⇒ P = 120000/6
∴ P = 20000

Question: A sum of money triples itself in 6 years at compound interest. In how many years will it amount to 27 times itself at the same rate of interest?

Solution: 
let,
the sum = P

ATQ,
3P = P(1 + r)⁶
⇒ (1 + r)⁶ = 3

Again,
let, sum will 27 times in n years

Then,
27P = P(1 + r)ⁿ
⇒ 27 = (1 + r)ⁿ
⇒ 3³ = (1 + r)ⁿ
⇒ {(1 + r)⁶}³ = (1 + r)ⁿ
⇒ (1 + r)¹⁸ = (1 + r)ⁿ
∴ n = 18

Question: Mina lives 3 km west of Lina's house. Rina lives 8 km north of Lina's house and 3 km west of Tina's house. What is the minimum distance from Mina's house to Tina's house?

Solution:

So, distance,
(ad)² = (cd + ce)² + (ae)²
⇒ (ad)² = (3 + 3)² + (8)²
⇒ (ad)² = 36 + 64
⇒ (ad)² = 100
∴ ad = 10

So the minimum distance from Mina's house to Tina's house = 10 km

Question: Simple interest on a certain sum of money for 3 years at 8% per annum is half the compound interest on Tk. 4000 for 2 years at 10% per annum. The sum placed on simple interest is

Solution:
Given,
principal P = Tk. 4000
Compound rate, r = 10% per annum
Time n = 2 years

Amount,
C = P(1 + r)ⁿ
= 4,000(1 + 10/100)²
= 4,000(11/10)²
= (4,000 × 121)/100
= 4840
∴ Compound interest =Tk. (4840 - 4000)
= Tk. 840

∴ Simple interest I = (840 ÷ 2) = Tk. 420
Rate r = 8%
Time n = 3 years

We know,
P = I/rn
= 420/{(8/100) × 3}
= (420 × 100)/(8 × 3)
= 1750

Question: Mr. Rony invested an amount of Tk. 1800 divided in two different schemes X and Y at the simple interest rate of 12% p.a. and 10% p.a. respectively. If the total amount of simple interest earned in 3 years be Tk. 558, what was the amount invested in Scheme Y?

Solution:
Let,
The amount invested in Scheme X = Tk. a
Then, the amount invested in Scheme Y = Tk. (1800 - a)

So the interest of X schemes =Tk. {a × (12/100) × 3}
= 9a/25
So the interest of Y schemes =Tk. {(1800- a) × (10/100) × 3}
= 540 - (3a/10)

ATQ,
(9a/25) + 540 - (3a/10) = 558
⇒ 9a/25 - 3a/10 = 558 - 540
⇒ (18a - 15a)/50 = 18
⇒ 3a/50 = 18
⇒ 3a = 18 × 50
⇒ 3a = 900
∴ a = 300

Amount in Scheme Y = Tk. (1800 - 300)
= Tk. 1500

Question: In how many years will the profit of Tk. 10,000 be Tk. 4,800 at the rate of 12% profit? 

Solution:
Given,
P = Tk. 10000
I = Tk. 4800
r = 12% = 12/100

We know,
n = I/Pr
= 4800/(10000 × 12/100)
= (4800 × 100)/(10000 × 12)
= 4 years

Question: How many degrees are between the hands of a clock at 4:30?

Solution: 
Value of angle = {(11×30) - (60 × 4)}/2
= (330 - 240)/2
= 90/2
= 45°

Question: For how much money will the profit at the rate of Tk. 5 per annum in 2 years 6 months be same as that of Tk. 1000 at the rate of Tk. 6 per annum in 4 years?

Solution:
Here are 2 parts,

In one part,
Given,
Principal P = Tk. 1000
Rate r = 6 per annum
Time n = 4 years

So simple interest I = (P × r × n)
= (1000 × 6/100 × 4)
= 240

In the other part,
I = Tk. 240
r = 5% = 1/20
n = 2 years 6 months
= 2.5 years

∴ P = I/rn
= 240/(1/20 × 2.5)
= (240 × 20)/(1 × 2.5)
= (4800 × 10)/25
= Tk. 1920

Question: Present population of a city is 90 lac. What will be the population of the city after 3 years if the growth rate of population of that city is 40 per thousand?

Solution:
Given,
Present population of the city is P = 90,00,000
growth rate of population, r = (40/1000) × 100%
= 4%
time n = 3 years

Here, in the case of growth of population, formula for compound principal is applicable
∴ C = P(1 + r)ⁿ
= 90,00,000(1 + 4/100)³
= 90,00,000(104/100)³
= 90,00,000 × (104/100) × (104/100) × (104/100)
= 9 x 104 x 104 x 104 
= 1,01,23,776

Question: If in a certain language PROPOSE is coded as PPONOQE, how is MENTORS coded in that code?

Solution:
In the given code
PROPOSE = PPONOQE

The 1^st, 3^rd 5^th and 7^th letters are same but in the place of 2^nd, 4^th and 6^th letters previous two letters are used.
So,
MENTORS = MCNROPS