ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Question: Iqbal invested an amount of Tk. 21250 for 6 years. At what rate of simple interest will be obtain the total amount of Tk. 26350 at the end of 6 years?

Solution:
Here, P = Tk. 21250,
Simple interest, I = Tk. (26350 - 21250) = Tk. 5100
Time, n = 6 years.

We know, 
I = Pnr
r = I/Pn
= {(5100 × 100)/(21250 × 6)}%
= 4%

Question: The compound interest on 300 tk. at 7% per annum is 21 taka. The period is:

Solution: 
চক্রবৃদ্ধি সুদাসল = P (1 + r)ⁿ
= 300 + 21
= 321 tk.

প্রশ্নমতে,
321 = 300 × (1 + 7/100)ⁿ
⇒ (1 + 7/100)ⁿ = 321/300
⇒  (107/100)ⁿ = 107/100
⇒  (107/100)ⁿ = (107/100)¹
∴ n = 1 year

Question: What number will replace the '?' mark?
2, 4, 12, 48, 240, ?, 10080

Solution:
২য়, ৩য়, ৪র্থ, ৫ম, ৬ষ্ঠ সংখ্যাকে যথাক্রমে 2, 3, 4, 5, 6, 7 দ্বারা ক্রমান্বয়ে গুণ করা হয়েছে।
2 × 2 = 4
4 × 3 = 12
12 × 4 = 48
48 × 5 = 240
240 × 6 = 1440
1440 × 7 = 10080

Question: Which one will sit in the blank?
QPO, NML, KJI, _____, EDC 

Solution:
এই সিরিজটি একটি বিপরীত বর্ণানুক্রমিক অক্ষর নিয়ে গঠিত।
শেষের দিক থেকে C D E,  F G H,  I J K,  L M N,  O P Q

∴  শূন্যস্থানে HGF বসবে।

Question: The population of a village decreases at the rate of 25% per annum. If its population 2 years ago was 24000, the present population is =?

Solution: 
Given that,
The population of a village 2 years ago = 24000
Rate of growth = 25% p.a.
Time period = 2 years

Present population = P{(1 - (r/100)}ⁿ
= 24000 × {1 - (25/100)}²
= 24000 × {1 - (1/4)}²
= 24000 × (3/4)²
= 24000 × (9/16)
= 13500

Question: A sum of Tk 7800 gives a simple interest of Tk. 702 in 2 years and 3 months. The rate of interest per annum is =?

Solution:
Time = 2 years 3 months
= 2 + (3/12)
= 2 + (1/4)
= (8 + 1)/4
= 9/4 years

Here,
I = 702
P = 7800
n = 9/4
r = ?

We know,
I = Pnr
⇒ r = I/pn
⇒ r = (702 × 4 × 100)/(7800 × 9)
∴ r = 4%

Question: A man starts from a point and travels 12 km in the north direction. Again he went east for 6 km and changed direction and again went north for 8 km. Finally went west for 6 km. What is the direct distance between his destination and journey?

Solution:
 
ঐ ব্যক্তির যাত্রাস্থান A এবং গন্তব্য স্থান E
∴ তার গন্তব্য স্থান ও যাত্রা স্থানের সরাসরি দূরত্ব, AE = (12 + 8) কি.মি.
= 20 কি.মি.

Question: A sum of money at compound interest doubles itself in 15 years. It will become eight times of itself in-

Solution: 
let the sum P 

2P = P (1 + r)¹⁵
⇒ (1 + r)¹⁵ = 2

let, sum will 8 times in n years
8P = P(1 + r)ⁿ
⇒ 8 = (1 + r)ⁿ
⇒ 2³ = (1 + r)ⁿ
⇒ {(1 + r)¹⁵}³ = (1 + r)ⁿ
⇒ (1 + r)⁴⁵ = (1 + r)ⁿ
∴ n = 45 years

Question: A sum was put at simple interest at a certain rate for 3 years. Had it been put at 1% higher rate, it would have fetched 510 taka more. The sum is - 

Solution: 
Let the sum be Tk x and original rate be R%

ATQ,
((x × (R +1) × 3)/100) - {(x × R × 3)/100} = 510
⇒ 3Rx + 3x - 3Rx = 51000
⇒ 3x = 51000
⇒ x = 17000

Hence, sum = Tk. 17,000

Question: Students of a class stand in a queue. If Raju is 22^nd in order from both ends, how many students are there in the queue?

Solution:
এক দিক থেকে রাজুর অবস্থান ২২ তম অর্থাৎ সে সহ ২২ জন।
অপর দিক থেকে তার অবস্থান ২২ তম অর্থাৎ সে বাদে আরও ২১ জন আছে।
∴ ঐ সারিতে মোট ছাত্র আছে (২২ + ২১) জন = ৪৩ জন 

Question: 4600 Taka is lent at compound interest of 10% per annum for 2 years. Find the amount after two years.

Solution:
Compound amount = P (1 + r)ⁿ
= 4600 {1 + (10/100)}²
= 4600 × (110/100) × (110/100)
= 46 × 11 × 11
= 5566

Question: A sum of Tk. 10,000 amounts to Tk. 12,000 in 4 years at the rate of simple interest. What is the rate of interest?

Solution: 
Interest = 12000 - 10000 = Tk. 2000

Let, interest rate = r%  

We know,
I = Pnr
⇒ 2000 = 10000 × 4 × r/100
⇒ r = 5

Interest rate 5%

Question: P is bigger than J. S is bigger than M. R is not as big as S but is bigger than J. M is not as big as J. Which is the smallest?

Solution: 
P is bigger the J. ⇒ P > J
S is bigger than M. ⇒ S > M
R is not as big as S but is bigger than J. ⇒ S > R > J
M is not as big as J. ⇒ J > M

∴ M is the smallest. 

Question: A sum of money amounts to Tk. 5200 in 5 years and to Tk. 5680 in 7 years at simple interest. The rate of interest per annum is?

Solution:
Simple interest for 2 years = (5680 - 5200) Tk.
= Tk. 480
∴ Simple Interest for 2 years =  Tk. 480
∴ Simple Interest for 5 years =  Tk. (480 × 5/2)
= Tk. 1200
∴ Principal = 5200 - 1200 = Tk. 4000.

We know, 
I = Pnr
⇒ r = I/Pn
⇒ r = (1200 × 100)/(4000 × 5)
∴ r = 6%

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 = Tk. 3225

Question: What number will replace the '?' mark?

Solution:
নিচের ১ম সংখ্যা + নিচের দুই সংখ্যার পার্থক্য = উপরের সংখ্যা।

১ম চিত্রে, 30 + (30 - 15) = 45
তৃতীয় চিত্রে, 28 + (28 - 21) = 35

∴ দ্বিতীয় চিত্রে, 25 + (25 - 19) = 31

Question: Which number will complete the series:
155, 151, 147, 143, 139, ........?

Solution:
The pattern is followed by
155 - 4 = 151
151 - 4 = 147
147 - 4 = 143
143 - 4 = 139
139 - 4 = 135

Hence the number= 135.

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