Bank Math

Question: A shopkeeper incurs a loss by selling an article for Tk 600. If he had sold it for Tk 900, he would have made a profit which is three times the initial loss. At what price should he sell the article to make 20% profit?

Solution:
ধরি, পণ্যের ক্রয়মূল্য = x টাকা
600 টাকায় বিক্রি করলে ক্ষতি = x - 600 টাকা
900 টাকায় বিক্রি করলে লাভ = 900 - x টাকা

প্রশ্নমতে,
900 - x = 3(x - 600)
⇒ 900 - x = 3x - 1800
⇒ 900 + 1800 = 3x + x
⇒ 2700 = 4x
∴ x = 675 টাকা

এখন, 20% লাভে,
ক্রয়মূল্য 100 টাকা হলে বিক্রয়মূল্য 120 টাকা
ক্রয়মূল্য 1 টাকা হলে বিক্রয়মূল্য 120/100 টাকা
∴ ক্রয়মূল্য 675 টাকা হলে বিক্রয়মূল্য = (120 × 675)/100 টাকা
= 810 টাকা

∴ বিক্রয়মূল্য: Tk. 810

Let C's age be x years.
Then,
B's age = 2x years.
A's age = (2x + 2) years.
(2x + 2) + 2x + x = 27
5x = 25
x = 5.
Hence,
B's age = 2x = 10 years.

1 / (1/p + 1/q - 1/r)
= 1 / (qr+pr-pq / pqr)
= pqr / (qr + pr - pq)

Question: Tamim and Shakib invest Tk. 15,000 and Tk. 25,000 respectively. Tamim, being the active partner, gets 10% of the total profit as salary. If the annual profit is Tk. 10,000, find Tamim’s total earnings.

Solution:
Total profit = 10,000 Taka.
Tamim gets 10% of total profit as salary.
Tamim’s salary = 10% of 10,000
= (10/100) × 10,000
= 1,000 Taka.

Remaining profit after salary = 10,000 - 1,000
= 9,000 Taka.

Divide remaining profit in ratio of their capitals
Tamim : Shakib = 15,000 : 25,000
= 3 : 5

Tamim’s share = [3/(3 + 5)] × 9,000 
= [3/8] × 9,000
= 3,375 Taka

Tamim’s total earning = Tamim’s salary + Tamim’s profit share
= 1,000 + 3,375
= 4,375 Taka

∴ Tamim's total earning = 4,375 Taka

Question: Working 4 hours a day, P can complete a work in 8 days and working 8 hours a day, Q can complete the same work in 4 days. Working 8 hours a day, they can jointly complete the work in-

Solution:
Working 4 hours a day, P can complete the work in 8 days
∴ P can complete the work in = 4 × 8 = 32 hours
∴ P can complete in 1 hour = 1/32 part

Working 8 hours a day, Q can complete the work in 4 days
∴ Q can complete the work in = 8 × 4 = 32 hours
∴ Q can complete in 1 hour = 1/32 part

(P + Q)'s 1 hour's work,
(1/32) + (1/32) = 2/32 = 1/16

∴ P and Q can complete the work in 16 hours
So, working 8 hours a day they require = 16/8
= 2 days

Question: An accurate clock shows 10:00 AM. Through how many degrees will the hour hand rotate when the clock shows 4:00 PM?

Solution: 
10:00 AM থেকে 4:00 PM পর্যন্ত অতিবাহিত সময় = 6 ঘণ্টা।

আমরা জানি,
ঘণ্টার কাঁটা 12 ঘণ্টায় ঘোরে 360°
∴ 1 ঘণ্টায় ঘোরে = 360°/12 = 30°
∴ 6 ঘণ্টায় ঘোরে = 6 × 30° = 180°

অতএব, 10:00 AM থেকে 4:00 PM পর্যন্ত ঘণ্টার কাঁটাটি 180° ঘুরবে।

Question: A bag contains 5 red balls, 7 blue balls and 6 green balls. One ball is drawn at random and replaced with 2 green balls. What is the probability that the first ball drawn was either red or blue and the second drawn was green in colour?

Solution:
Number of Red balls = 5  
Number of Blue balls = 7  
Number of Green balls = 6  

Total number of balls = 5 + 7 + 6 = 18  

After the first ball is drawn, it is replaced with 2 green balls.
So the total number of balls becomes 18 - 1 + 2 = 19,
and the number of green balls becomes 6 + 2 = 8.

So, Required probability  
= (5/18) × (8/19) + (7/18) × (8/19)
= 8/19 [ (5/18) + (7/18)]
= (8/19) × (12/18)
= 16/57

Question: In a class of 50 students, 18 students like Biology, 20 students like Chemistry, and 22 students like Physics. It is found that 4 students like both Biology and Chemistry, 5 students like both Biology and Physics, and 6 students like both Chemistry and Physics. If 3 students like none of these subjects, find the number of students who like all three subjects.

Solution:
Let the number of students who like all three subjects = x.
Students liking at least one subject = Total students - Students liking none
= 50 - 3
= 47

According to question,
⇒ Biology + Chemistry + Physics - (Bio & Chem + Bio & Phys + Chem & Phys) + x = 47
⇒ 18 + 20 + 22 - (4 + 5 + 6) + x = 47
⇒ 60 - 15 + x = 47
⇒ 45 + x = 47
∴ x = 2

Question: What is the value of sin60°

Solution:
 sin60° = √3/2
sin30° = 1/2
sin45° = 1/√2
tan30° = 1/√3

Question: When a positive integer X is divided by Y, the quotient is 11 and the remainder is 5. When X is divided by (Y + 3) the quotient is 9 and the remainder is 2. What is the value of X?

Solution: 
Given that,
When X is divided by Y. Then we get,
X = 11Y + 5 .......(1) 

And, 
When X is divided by Y + 3. Then we get,
X = 9(Y + 3) + 2 = 9Y + 27 + 2 = 9Y + 29.......(2) 

From (1) and (2), Then we get,
⇒ 11Y + 5 = 9Y + 29 
⇒ 11Y - 9Y = 29 - 5
⇒ 2Y = 24
⇒ Y = 24/2
∴ Y = 12

From (1),
X = 11Y + 5 = (11 × 12) + 5 = 132 + 5 = 137

So the value of X is 137. 

Question: A geometric series has its first term as 1 divided by square root of 2, and its common ratio is √2. Which term in the sequence is 16√2?

Solution: 
First term, a = 1/√2
Common ratio, r = √2

Let, the n-th term be = ar^{n - 1} = 16√2
⇒ (1/√2) (√2)^{n - 1} = 16√2
⇒ (√2)^{n - 1} = 32
⇒ (√2)^{n - 1} = (√2)¹⁰
⇒ n - 1 = 10 
∴ n = 11

So the 11^th term is 16√2.

Number of shares sold = 7500/100
= 75

Proceeds from sale of Tk. 7500 stock
= Tk. [(105.50 - 0.5) × 75]
= Tk. 7875

Number of new shares purchased
= 7875/)124.50 + 0.50
= 7875/125
= 63

Original income
= 10% of Tk. 7500
= Tk. 750

New income
= 14% of Tk. 6300
= Tk. {(14/100) × 6300}
= Tk. 882

∴ Change in income
= Tk. (882 - 750)
= Tk. 132

Question: The ratio of milk and water in a solution is 7 : 4. After adding 8 liters of water, the ratio of milk and water becomes 3 : 2. Find the final amount of water in the solution.

Solution:
Let the initial amount of milk = 7x liters
Let the initial amount of water = 4x liters

According to the question,
7x/(4x + 8) = 3/2
⇒ 2 × 7x = 3 × (4x + 8)
⇒ 14x = 12x + 24
⇒ 14x - 12x = 24
⇒ 2x = 24
⇒ x = 12

∴ Final amount of water = 4x + 8
= 4 × 12 + 8
= 48 + 8
= 56 liters

Question: The area of a trapezium is 120 square cm. The length of one of the parallel sides is 10 cm, and the distance between the parallel sides is 15 cm. Find the length of the other parallel side.

Solution:
দেওয়া আছে,
ট্রাপিজিয়ামের ক্ষেত্রফল = 120 সেমি²
একটি সমান্তরাল বাহু, a = 10 সেমি
সমান্তরাল বাহুদ্বয়ের মধ্যবর্তী দূরত্ব, h = 15 সেমি

ধরি, অপর সমান্তরাল বাহু = b সেমি

আমরা জানি,
ট্রাপিজিয়ামের ক্ষেত্রফল = (1/2) × (a + b) × h

প্রশ্নমতে,
120 = (1/2) × (10 + b) × 15
⇒ 120 × 2 = (10 + b) × 15
⇒ 240 = (10 + b) × 15
⇒ (10 + b) = 240 / 15
⇒ 10 + b = 16
⇒ b = 16 - 10
⇒ b = 6 সেমি

সুতরাং, অপর সমান্তরাল বাহুটির দৈর্ঘ্য 6 সেমি।

Question: Rahim ran half the distance at 5 km/h and the remaining half at 10 km/h. What was his average speed for the entire run?

Solution:
Let the total distance = 2x km, so each half = x km.

Time for first half = x / 5 hours
Time for second half = x / 10 hours

∴ Total time = (x/5) + (x/10) = (2x + x)/10 = 3x/10 hours

∴Average speed = Total distance / Total time
= 2x ÷ (3x/10)
= 2x × (10/3x)
= 20/3
= 6.67 km/h 

Let the price paid by Prateek be Tk. x
Then,
140% of 120% of x=10500
⇒(140/100)×(120/100)×x=10500
⇒ x=(10500×25/42)=6250

Let us consider the options
a) 11 - 2 = 9 and 9/3 = 3; 11 - 4 = 7 and 7/7 = 1
b) 22 - 2 = 20 but 20/3 = 6.67
c) 18 - 2 = 16 but 16/3 = 5.33
d) 32 - 2 = 30 and 30/3 = 10; 32 - 4 = 28 and 28/7 = 4
As the requirement is the least possible value, answer will be 11

Question: Which number replaces the question mark?


Solution:
প্রথম কলামের ক্ষেত্রে,
27 - (7 × 2) = 27 - 14 = 13

দ্বিতীয় কলামের ক্ষেত্রে,
144 - (45 × 2) = 144 - 90 = 54

একইভাবে, তৃতীয় কলামের ক্ষেত্রে,
68 - (32 × 2) = 68 - 64 = 4

∴ প্রশ্নবোধক চিহ্নের স্থানে 4 বসবে।

The series is: 3 + 0² = 3, 3 + 1² = 4, 4 + 2² = 8, 8 + 3² = 17, 17 + 4³ = 33, 33 + 5² = 58    

Question: If two numbers are respectively 30% and 40% more than a third number, what percent is the first of the second?

Solution:
Let the third number be 100
Then,
1st number = 100 + 30 = 130
2nd number = 100 + 40 = 140

To find what percent the first number is of the second number is,
=(130 × 100)/140
= 650/7
=92.86%

∴ The first number is 92.86% of the second number.

Question: If the sum of three consecutive odd integers is 117, what is the largest number?

Solution:
Let the three consecutive odd integers be:
x, x + 2, x + 4

Then:
x + (x + 2) + (x + 4) = 117
⇒ 3x + 6 = 117
⇒ 3x = 117 − 6
⇒ 3x = 111
⇒ x = 111 ÷ 3
⇒ x = 37

So the numbers are: 37, 39, 41 (Largest)

∴ The largest number is 41.

Question: Yesterday, a bookstore had Tk. 3,600 in total sales. If 1/4 of the total sales were from the sale of books that had been reduced by 25 percent, what would have been the total sales yesterday if all of the books had been sold at full price?

Solution:
মোট বিক্রয় = Tk. 3,600

ডিসকাউন্ট দেওয়া বইয়ের বিক্রয়:
= Tk. 3,600 × (1/4)
= Tk. 900

এই Tk. 900 এসেছে 25% ছাড়ের পরে বিক্রি থেকে, অর্থাৎ এটি আসল মূল্যের 75%।
∴ আসল মূল্য = 900 ÷ 0.75 = Tk. 1,200

বাকি বই পূর্ণ দামে বিক্রি হয়েছে:
= মোট বিক্রি - ডিসকাউন্ট বই বিক্রি
= Tk. 3,600 - Tk. 900
= Tk. 2,700

∴ সব বই পূর্ণ দামে বিক্রি হলে মোট বিক্রয় হতো:
= Tk. 1,200 + Tk. 2,700
= Tk. 3,900

Question: A dishonest dealer marks up the price of his goods by 20% and gives a discount of 10% to the customer. He also uses a 900 gram weight instead of a 1 kilogram weight. Find his percentage profit due to these maneuvers?

Solution:
Let Cost Price, CP = Tk. 1 per gram

Dealer marks up by 20%,
Then the Marked Price of 1000 gram is = 1000 + 20% of 1000
= 1000 + 200 =Tk. 1200

Now Dealer gives 10% discount,
So, Selling price after discount = 1200 - 10% of 1200
= 1200 - 120 = Tk. 1080

Then, Dealer is dishonest and sells 900 grams for the price of 1080 Tk.
The Cost Price of 900 grams is Tk. 900 (since the cost price per gram is Tk. 1).

Profit = 1080 - 900 = 180
so, Percentage profit = (Profit/Cost Price) × 100%
= (180/900) × 100%
= 20%

Speed of the man in still water = 12 km/hr.
Speed of the stream = 3 km/hr.
Speed downstream = 12 + 3
= 15 km/hr.
Speed upstream = (12 - 3)
= 9 km/hr.
Average speed = (Speed downstream × Speed upstream)/Speed in still water
= (15 × 9)/12
= (15 × 3)/4
= 45/4
= 11(1/4) km/hr.

Question: Find the value of x if log_x 256 = 4.

Solution:
log_x 256 = 4
⇒ x⁴ = 256
⇒ (x²)² = 256
⇒ x² = 16
⇒ x = √16
⇒ x = 4

∴ The value of x is 4.

Question: The perimeter of a rectangular field is 80 meters. If the ratio of its length to its width is 5:3, what is the area of the field in square meters?

Solution:
ধরি,
আয়তাকার ক্ষেত্রের দৈর্ঘ্য = 5x মিটার এবং প্রস্থ = 3x মিটার।
পরিসীমা, P = 80 মিটার

আমরা জানি,
আয়তক্ষেত্রের পরিসীমা, P = 2 × (দৈর্ঘ্য + প্রস্থ)
∴ 80 = 2 × (5x + 3x)
⇒ 80 = 2 × (8x)
⇒ 80 = 16x
⇒ x = 80/16
∴ x = 5

সুতরাং,
দৈর্ঘ্য = 5x = 5 × 5 = 25 মিটার
প্রস্থ = 3x = 3 × 5 = 15 মিটার

∴ ক্ষেত্রফল = দৈর্ঘ্য × প্রস্থ
= 25 × 15 বর্গ মিটার
= 375 বর্গ মিটার

অতএব, মাঠটির ক্ষেত্রফল = 375 বর্গ মিটার।

Question: A hollow cylinder has an internal radius of 6 cm, an external radius of 10 cm, and a height of 7 cm. Find the volume of the material used to make the cylinder.

Solution: 
Let, 
Internal radius (r) = 6 cm
External radius (R) = 10 cm
Height (h) = 7 cm

Volume of the the material used, 
V = πh(R² - r²) 
= π × 7 (100 - 36)
= π × 7 × 64
= 448π

Let,
Initial milk = 4x
Initial water = 3x
ATQ,
(3x + 13)/4x = 1/1
Or, 4x = 3x + 13
Or, x = 13
So, water at final = (3 × 13) + 13
= 52 L

Question: Two dice are thrown simultaneously. What is the probability of getting the face numbers are same?

Solution:
In a simultaneous throw of two dice, we have n(S) = 6 × 6 = 36
Let E = event of getting two numbers are same.
Then E = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}
therefore, n(E) = 6
And P(E) = P (getting two numbers are same)

∴ P(E) = n(E)/n(S)
= 6/36
= 1/6

Hence the answer is 1/6 .

Question: Father is aged three times more than his son Rakib. After 8 years, he would be two and a half times of Rakib's age. After further 8 years, how many times would he be of Rakib's age? 

Solution:
Let,
Rakib's present age be x years.
Then, father's present age =(x + 3x) years
= 4x years.

ATQ,
(4x + 8) = {5(x + 8)}/2
⇒ 8x + 16 = 5x + 40
⇒ 3x = 24
∴ x = 8

Now,
(4x + 16)/(x + 16)  = 48/24 = 2

∴ Father would be 2 times of Rakib's age.

Question: What is the H.C.F. of the following fractions?
6/10, 9/15, 12/20.

Solution:
আমরা জানি,
ভগ্নাংশের গসাগু = লবের গসাগু/হরের লসাগু

এখানে,
লব = 6, 9, 12
6 = 2 × 3
9 = 3 × 3
12 = 2 × 2 × 3

∴ গসাগু (HCF) = 3

হর = 10, 15, 20
10 = 2 × 5
15 = 3 × 5
20 = 2 × 2 × 5

∴ লসাগু (LCM) = 2 × 2 × 3 × 5
= 60

ভগ্নাংশের গসাগু = লবের গসাগু/হরের লসাগু
= 3/60
=1/20

Let the sum be x Assume that it will treble in n years.
Note that when the money doubles, simple interest is (2x - x), and when the money trebles, simple interest is (3x - x)
simple interest ∝ T (because here P and R are constants)
Therefore,
(2x - x) : (3x - x) = 12 : n
⇒ x : 2x = 12 : n
⇒ n = 24.

Question: A tank is 1/3 parts full with water. If 16 liters of water is added, the tank becomes 5/6 parts full. What is the capacity of the tank?
(Janata RC 2022 অনুযায়ী)

Solution:
ধরি,
ট্যাংকের ধারণক্ষমতা = x লিটার

প্রশ্নমতে,
(x/3) + 16 = 5x/6
⇒ (5x/6) - (x/3) = 16
⇒ (5x - 2x)/6 = 16
⇒ 3x = 96
⇒ x = 96/3
⇒ x = 32

অর্থাৎ ট্যাংকের ধারণক্ষমতা = 32 লিটার 

Question: A lift can carry 12 adults or 20 children at a time. If there are 9 adults in the lift, how many children can be loaded onto it?

Solution:
এখানে,
12 জন প্রাপ্তবয়স্ক (Adults) = 20 জন শিশু (Children)
∴ 1 জন প্রাপ্তবয়স্ক = 20 / 12 = 5/3 জন শিশু
∴ 9 জন প্রাপ্তবয়স্ক = 9 × 5/3 = 15 জন শিশু

লিফটের সর্বোচ্চ ধারণক্ষমতা = 20 জন শিশু
∴ আরও শিশু বহন করা যাবে = 20 - 15 = 5 জন শিশু

সুতরাং, 9 জন প্রাপ্তবয়স্কের সাথে আরও 5 জন শিশুকে লিফটে নেওয়া যাবে।

Question: The average temperature on Monday, Tuesday, and Wednesday was 26°C. The average temperature on Tuesday, Wednesday, and Thursday was 25°C. If the temperature on Monday was 28°C, what was the temperature on Thursday?

Solution:
The total temperature on Monday, Tuesday, and Wednesday = 26° × 3 = 78°
The total temperature on Tuesday, Wednesday, and Thursday = 25° × 3 = 75°
 
ATQ,
(Mon + Tue + Wed) - (Tue + Wed + Thu) = 78° - 75°
⇒ Mon - Thu = 3°
⇒ Thu = Mon - 3°
⇒ Thu = 28° - 3°
∴ Thu = 25°

∴ The temperature on Thursday = 25°

Distance covered = 120 + 120 = 240 m
Time = 12s
Let the speed of each train = v.
Then relative speed = v + v = 2v
2v = distance/time
= 240/12
= 20 m/s
Speed of each train = v = 20/2
= 10 m/s
= 10 × 36/10 km/hr
= 36 km/hr

Question: If (3 + √p) > 2√p, which of these statements cannot be false?

Solution:
(3 + √p) > 2√p
⇒ 3 > 2√p - √p
⇒ 3 > √p
⇒ √p < 3
⇒ p < 3²
∴ p < 9

যেহেতু,
N ঋনাত্মক পূর্ণসংখ্যা।
ধরি, N = -7
∴ (N)² = (-7)² = 49
6 - N = 6 - (-7) = 13
- N = -(-7) = 7
6 + N = 6 + (-7) = -1
∴ 6 + N এর মান ঋনাত্মক।

Ratio of the initial capital of A and B = 4 : 5
Hence we can take the initial capitals of A and B as 4x and 5x respectively.
Ratio in which profit will be divided
According to the question,
= (4x × 3) + {3/4 × (4x × 7)} : (5x × 3) + {4/5 × (5x × 7)}
= (12 + 21) : (15 + 28)
= 33 : 43
A's share = 760 × (33/76)
= 330.

Suppose, Cost Price = x & Profit = y
So, x + y = 30,000.......... (i)
x + 0.9y = 29,500....... (ii)
(i) - (ii), we get
y = 5000
So, Cost Price = 30,000 - 5000 = 25,000 Taka
বিকল্প পদ্ধতিঃ
লাভ কমলো = ৩০,০০০ - ২৯,৫০০ = ৫০০ টাকা।
প্রশ্নমতে,
১০ % = ৫০০
∴ ১০০ % = (৫০০ × ১০০)/১০ = ৫০০০ টাকা।
অর্থাৎ,
বিক্রয়মূল্য = ৩০, ০০০ টাকা।
মোট লাভ = ৫,০০০ টাকা।
∴ ক্রয়মূল্য = (৩০,০০০ - ৫,০০০) টাকা।
= ২৫,০০০ টাকা।

Since face value is not given, take it as Tk. 100.
As it is an 8% stock, income (dividend) per stock = Tk. 8
ie, For an income of Tk. 8, amount of stock needed = Tk. 100
For an income of Tk. 800, the amount of stock needed = (100 × 800)/8
= 10000

Question: If the chairperson's seat is fixed, in how many ways can 6 people be seated at a circular table?

Solution:
Chairperson’s seat is fixed
∴ Remaining 5 people can be arranged in 5! ways
= 5 × 4 × 3 × 2 
= 120 ways

Question: The area of the largest rectangle that can be inscribed in a circle of radius r is: 

Solution:
The largest rectangle inscribed in a circle is a square.
Diagonal of the square = diameter of the circle = 2r.

Let, side of square = s
By Pythagoras:
s² + s² = (2r)²
⇒ 2s² = 4r²
⇒ s² = 2r²
⇒ s = r√2

∴ Area = s² = 2r²

Question: A man sells 2 commodities for Tk. 3500 each, neither losing nor gaining in the deal. If he sold one commodity at a gain of 25%, then what is the cost price of another commodity?

Solution: 
Total S.P. = Tk. 7000 and
Total C.P. = Tk. 7000
Let,
S.P. of 1st commodity = Tk. 3500
Gain on it = 25%

∴ C.P. of 1st commodity
= Tk. {(100/125)×3500}
= Tk. 2800

C.P. of 2nd commodity = Tk. (7000 - 2800) = TK. 4200