ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Question: A sum of money amounts to Tk. 21000 in 2 years at 20% simple interest per annum. Find the sum.

Solution: 
Here,
A = 21000, T = 2, R = 20%
⇒ A = P + SI
⇒ A = P + (P × R × T/100)
⇒ A = P [1 + (R × T/100)]
⇒ 21000 = P [1 + 0.4]
⇒ P = 21000/1.4
⇒ P = 15000
Thus, the required sum is Tk. 15000

Question: Find the amount if Tk. 20000 is invested at 10% compound interest p.a. for 3 years.

Solution: 
Using the formula:
C = P [1 + R/100]ⁿ
C = 20000 [1 + (10/100)]³
= 20000 × 1.1 × 1.1× 1.1
= 26620

Question: There are 120 chickens and goats on a farm. There are 70 more chickens than goats. How many goats are there?

Solution:
Let,
The number of goats be x
∴ The number of chickens x + 70

ATQ,
x + x + 70 = 120
⇒ 2x = 50
∴ x = 25

Question: A man borrowed a certain sum of money at the rate of 6% per annum for the first two years, 9% per annum for the next three years, and 14% per annum for the period beyond 5 years. If he pays a total interest of Tk. 20900 at the end of 9 years, find the amount he borrowed.

Solution: 
Let the borrowed sum be P.
SI for first 2 years + SI for next 3 years + SI for next 4 years = 20900
⇒ (P × 6 × 2/100) + (P × 9 × 3/100) + (P × 14 × 4/100) = 20900
⇒ (95P)/100 = 20900
⇒ P = (20900 × 100)/95
⇒ P = 22000
Therefore, Borrowed sum = Tk. 22000

Question: Find the compound interest, if Tk. 1000 was invested for 1.5 years at 20% p.a. compounded half yearly.

Solution:
 As it is said that the interest is compounded half yearly. So, the rate of interest will be halved and time will be doubled.
CI = P [1 + (R/100)]ⁿ - P
CI = 1000[1 +(10/100)]³ - 1000
CI = 1000 × 1.1 × 1.1 × 1.1 - 1000
CI = 1331 - 1000
CI = 331

Question: If all Bloops are Razzies and all Razzies are Lazzies, then ___.

Solution:
From the information we get,

∴ All Bloops are definitely Lazzies.

Question: At what annual rate of interest will a sum of money be thrice in 10 years?

Solution: 
Amount = Principal + SI
If the sum of money would be thrice the principal after 10 years, the SI would be twice the principal.
⇒ SI = 2 × P
⇒ (P × R × T / 100) = 2 × P
⇒ R × T / 100 = 2
⇒ R × T = 200
⇒ R × 10 = 200
⇒ R = 20%
Thus, the required rate of interest is 20%

Question: The compound interest on a sum of Tk. 625 in 2 years is Tk. 51. Find the rate of interest.

Solution: 
We know that A = CI + P
A = 51 + 625 = 676

Now going by the formula:
A = P[1+(R/100)]ⁿ
⇒ 676 = 625[1 + (R/100)]²
⇒ 676/625 = [1 + (R/100)]²
We can see that 676 is the square of 26 and 625 is the square of 25
Therefore,
⇒ (26/25)² = [1 + (R/100)]²
⇒ 26/25 = [1 + (R/100)]
⇒ 26/25 - 1 = R/100
⇒ 1/25 = R/100
⇒ R = 100/25
∴ R = 4

Question: 13, 17, 19, 23, 29, ?
Which number replace the Question mark?

Solution:
13, 17, 19, 23, 29 
All the numbers are prime number
∴ Next number will be 31

Question: The simple interest on a sum of money in 5 years at 12% per annum is Tk. 500 less than the simple interest accrued on the same sum in 7 years at 10% per annum. Find the sum.

Solution: 
Let the sum be P.
SI in 5 years at 12% per annum = P × 12 × (5/100) = 0.6 P
SI in 7 years at 10% per annum = P × 10 × (7/100) = 0.7 P

Now, according to the question,
0.7 P - 0.6 P = 500
⇒ 0.1 P = 500
⇒ P = 5000
Thus, the required sum is Tk. 5000

Question: A sum of money is put on compound interest for 2 years at 20%. It would fetch Tk. 482 more if the interest is payable half yearly than if it were payable yearly. Find the sum.

Solution: 
Let the Principal = Tk. 100
When compounded annually,
A = 100 [1 + 20/100]²
= 100 × 1.2 × 1.2 
= 144

When compounded half yearly,
A = 100[1 + 10/100]⁴
= 100 × 1.1 × 1.1 × 1.1 × 1.1
= 146.41

Difference, 146.41 - 144 = 2.41

If difference is 2.41, then Principal = Tk. 100
If difference is 482, then Principal = (100/2.41) × 482 = 20000

Question: Which fraction is the biggest?

Solution:
3/5 = 0.6
5/8 = 0.625
1/2 = 0.5
4/7 = 0.571

Question: A sum of Tk. 1000 was lent to two people, one at the rate of 5% and the other at the rate of 8%. If the simple interest after one year is Tk. 62, find the sum lent at 5% rate.

Solution: 
Let the sum lent at 5% be P.
∴ Sum lent at 8 % = 1000 - P

Now, according to the question,
SI for 5% + SI for 8% = 62
⇒ (P × 5 × 1/100) + ((1000 - P) × 8 × 1/100) = 62
⇒ 5P + 8(1000 - P) = 6200
⇒ 5P + 8000 - 8P = 6200
⇒ 3P = 1800
⇒ P = 600
Therefore, sum lent at 5 % = P = Tk. 600

Question: Lex has Tk. 1780.80 in his savings account that he opened 6 years ago. His account has an annual interest rate of 6.8% compounded annually. How much money did Lex use to open his savings account?

Solution:
We know that,
A = P[1+(R/100)]ⁿ
⇒ 1780.80 = P[1 + 6.8/100]⁶
⇒ 1780.80 = P[1 + 0.068]⁶
⇒ 1780.80 = 1.484P
⇒ P = 1780.80/1.484
∴ P = 1200

Question: How many triangles are there in the following figure.

Solution:

△AEF, △BEP, △EOP, △BPD, △OPD, △ODQ, △OQF, △CFQ, △CDQ, △BEO, △BED, △BOD, △DEO, △CFO, △CDO, △CFD, △DFO, △BCO, △DEF and △ABC
There are 20 triangles.

Question: What would be the annual interest accrued on a deposit of Tk. 10000 in a bank that pays a 4.5% per annum rate of simple interest?

Solution: 
Here,
P = 10000,
R = 4.5%,
T = 1

∴ SI = P × R × T/100
⇒ SI = 10000 × 4.5 × 1/100
⇒ SI = 450
Thus, the annual interest would be Tk. 450

Question: An amount of Tk. 25000 in 2 years at compound interest compounded annually, if the interest rate for the successive years be 4% and 5% per annum respectively, is-

Solution:
Principal (P) = Tk. 25000
Interest rates:
4% for the 1st year
and 5% for the 2nd year
Time (t) = 2 years

Amount for the first year: A₁ = P(1 + r₁/100)
Amount for the second year: A₂ = A₁(1 + r₂/100)

A₁ = 25000(1 + 4/100)
⇒ A₁ = 25000 × 1.04
⇒ A₁ = 26000

A₂ = 26000(1 + 5/100)
⇒ A₂ = 26000 × 1.05
⇒ A₂ = 27300

∴ The correct answer is option 1.

Question: What is the odd one out?

Solution:
Cotton, linen, and silk are all natural fibers derived from plants or animals.

Cotton comes from the cotton plant.
Linen comes from the flax plant.
Silk comes from silkworms.

Nylon, on the other hand, is a synthetic fiber made from petrochemicals, meaning it is man-made rather than derived from natural sources.
Thus, nylon is the odd one out.

Question: A sum of money becomes 6/5 of itself in 3 years at a certain rate of simple interest. The rate of interest per annum is-

Solution:
Let principal = 5P 
Hence amount = 5P × 6/5 = 6P 
SI = 6P - 5P = P 
Time = 3 years 

SI = P × R × T/100 
⇒ P = 5P × R × 3/100 
⇒ R = 100/15 
∴ R = 6.67% 

Question: Find compound amount and compound interest for Tk. 10000 at 3% interest per annum in 2 years.

Solution:
Principal (P) = Tk. 10000
Rate (R) = 3%
Time (n) = 2 years

A = P[1+(R/100)]ⁿ
A = 10000(1 + 3/100)²
= 10000 × 1.03 × 1.03
= 10609

Compound Amount (A) = Tk. 10609

Compound Interest (CI) = A - P
⇒ CI = 10609 - 10000
⇒ CI = 609

Question: Marathon is to race as hibernation is to-

Solution:
A marathon is a long race and hibernation is a lengthy period of sleep.
The answer is not choice a or b because even though a bear and winter are related to hibernation, neither completes the analogy.
(Choice c) is incorrect because sleep and dream are not synonymous.

Question: What amount of money will amount to 750 in 6 years and to 800 in 8 years at simple interest?

Solution:
In two years the interest becomes = 800 - 750 = 50Tk.
∴ Interest in 1 year = 25 
∴ Interest in 6 year = 25 × 6 = 150 

Principal = Amount - SI 
= 750 - 150 = Tk. 600 

Question: In how many years will Tk. 20 amounts to Tk. 24.20 at 10% per annum compound interest?

Solution:
Principal (P) = Tk. 20
Amount (A) = Tk 24.20
Rate (r) = 10% per annum
t = time in years

A = P(1 + r/100)^t 
24.20 = 20(1 + 10/100)^t
⇒ 24.20 = 20(1.1^)t
⇒ 24.20/20 = (1.1)^t
⇒ 1.21 = (1.1)^t
⇒ (1.1)² = (1.1)^t
∴ t = 2

Question: SCD, TEF, UGH, ____, WKL.

Solution:
There are two alphabetical series here.
The first series is with the first letters only: S, T, U, V, W.
The second series involves the remaining letters: CD, EF, GH, IJ, KL.

∴ Answer will be VIJ

Question: Simple Interest on certain sum of money of 10 year is 3130 but if after 5 year principal become 5 times then find the total SI after 10 years.

Solution:
SI of 10 years = 3130 
Simple Interest same in every year. 
So, SI in first 5 year = (3130/10) × 5 = 1565 

If principal become 5 times in 5 year than interest will also become 5 times. 
So, Interest of next 5 years= 1565 × 5 = 7825 
Total Simple Interest = 7825 + 1565 = 9390