Bank Math

Question: What is the probability of getting a sum of 9 when a die is thrown twice?

Solution: 
In two throws of a dice, n(S) = (6 x 6) = 36
Let E = the event of getting a sum
= {(3, 6), (4, 5), (5, 4), (6, 3)}
∴ n(E) = 4

Hence, P(E) = n(E)/n(S)
= 4/36
= 1/9

Question: There are 10 true-false questions in an examination, these questions can be answered in-
(পরীক্ষায় ১০টি সত্য-মিথ্যা প্রশ্ন আছে, এই প্রশ্নগুলোর উত্তর দেওয়া যেতে পারে কত উপায়ে?)

Solution: 
Total number of question = 10 
Each question has 2 answer.

∴ These question can be answered in 2¹⁰ ways = 1024 ways

perimeter of the plot = 2(90+50) = 280m
number of poles =280/5 = 56m

ধরি,
AB হচ্ছে দেয়াল এবং BC হচ্ছে মই।
এখানে ∠ACB = 60° এবং AC = 4.6 মিটার
তাহলে, AC/BC = cos 60° = 1/2 [ যেহেতু, cosθ = ভূমি/অতিভুজ]
⇒ BC = 2 × AC
∴ BC = 2 × 4.6
= 9.2m

এখানে,
30 দ্বারা বিভাজ্য সংখ্যা = 1000/30
ভাগফল 33 এবং ভাগশেষ 10
সুতরাং 30 দ্বারা বিভাজ্য সংখ্যা = 33 টি
30 ও 16 এর ল, সা, গু = 240
এখন 1000/240 =
ভাগফল 4 এবং ভাগশেষ 40
∴ 30 ও 16 দ্বারা বিভাজ্য সংখ্যা = 4 টি
সুতরাং, (33 - 4) = 29 টি সংখ্যা 30 দ্বারা বিভাজ্য কিন্তু 16 দ্বারা বিভাজ্য নয়।

Question: A team of 4 men and 3 women is to be formed from 6 men and 5 women. In how many ways can the team be formed?

Solution:
We have 6 men and 5 women.
We need to choose 4 men from 6 and 3 women from 5.

Number of ways to choose 4 men from 6:
⁶C₄ = 6!/(4!(6 - 4)!) 
= (6 × 5)/(2 × 1)
= 15

Number of ways to choose 3 women from 5:
⁵C₃ = 5!/(3!(5 - 3)!)
= (5 × 4)/(2 × 1)
= 10

Total number of ways to form the team = 15 × 10 = 150

Therefore, the team can be formed in 150 different ways.

Question: In an examination, 32% is the pass mark. If an examinee gets 14 marks and fails by 10 marks, what is the maximum mark?

Solution: 
If the candidate fails by 10 marks, it means the pass marks = 14 + 10
= 24

Let the maximum marks be x

ATQ,
32% of x = 24
⇒ x × (32/100) = 24
⇒ 32x = 2400
⇒ x = 2400/32
⇒ x = 75

Question: X buys a product for Tk. 400 and sells it to Y at a profit of 30%. Y then sells it to Z at a profit of 15%. How much does Z pay to Y?

সমাধান:
X এর 30% লাভে বিক্রয়মূল্য = 400 + 400 এর 30%
= 400 + (400 × 30/100)
= 400 + 120
= 520

X এর বিক্রয়মূল্য = Y এর ক্রয়মূল্য

Y এর 15% লাভে বিক্রয়মূল্য = 520 + 520 এর 15%
= 520 + (520 × 15/100)
= 520 + 78
= 598

সুতরাং, Y এর বিক্রয়মূল্য = Z এর ক্রয়মূল্য = Tk. 598

Question: If two men or four women or six boys can finish a job in 72 days, then find out the number of days in which that work can be done by one man, two women, and three boys?

Solution:
Given,
2 men = 4 women = 6 boys
1 man = 2 women = 3 boys
∴ 1 man + 2 women + 3 boys = (1 + 1 + 1) men
= 3 men

Here,
2 men can do in = 72 days
∴ 1 men can do in = (72 × 2) days
= 144 days
∴ 3 men can do in = (144 ÷ 3) days
= 48 days

Question: If tanθ = 5/12, then cosecθ = ?

Solution:
এখানে,
tanθ = 5/12 = লম্ব/ভূমি

∴ লম্ব = 5, ভূমি = 12

∴ অতিভুজ = √(5² + 12²)
= √(25+144) = √169 = 13

∴ cosecθ
= অতিভুজ/লম্ব
= 13/5

Let length of the train be x m
Speed of train,
(x+162) / 18 = (x+120) / 15
∴ x = 90 m

Question: Over a span of 9 hours, a man traveled 61 km, part of it on foot at 4 km/hr and part by bike at 9 km/hr. What distance did he walk?

Solution:
Let the time in which he travelled on foot = x hr
Then the time in which he travelled on bicycle =(9 - x) hr
distance = speed × time
⇒ 4x + 9(9 - x) = 61
⇒ 4x + 81 - 9x = 61
⇒ 5x = 20
⇒ x = 4

∴ The distance travelled on foot = 4 × 4 = 16 km

Question: In a class of 50 students, 20 study History, 25 study Geography, and 10 study both subjects. How many students study neither subject?

Solution:
Given that,
Total students = 50
Study History = 20
Study Geography = 25
Study both = 10

Students studying at least one subject = H + G - Both
= 20 + 25 - 10
= 35

Therefore, Students who study neither subject = Total students - at least one subject
= 50 - 35 = 15

∴ 15 students study neither subject.

Question: A right triangle has sides 9 cm, 12 cm, and 15 cm. What is its area?

Solution:
A right triangle with sides 9 cm, 12 cm, and 15 cm.

We know,
Area = (1/2) × base × height
= (1/2) × 9 × 12
= 54 cm²

So, the area of the right triangle is 54 cm².

দেওয়া আছে,
4^{x + 1} = 32
⇒ 2^{2(x + 1)} = 2⁵
⇒ 2(x + 1) = 5
⇒ 2x + 2 = 5
⇒ 2x = 5 - 2
⇒ 2x = 3
∴ x = 3/2

If AB = BC, then ∠A =  ∠C
As we know, ∠A +  ∠B + ∠C = 180°
Or, A +  70° + ∠A = 180°
Or, 2∠A = 180° - 70° = 110°
So, ∠A = 55°

Question: In a business partnership, Ajay and Vijay invested Tk. (1400 + x) and Tk. (1800 + 2x). After 12 months, Vijay’s profit amounted to Tk. 3200 out of a total Tk. 5600. Calculate Ajay’s starting investment.

Solution:
Ratio of profit share = Ajay : Vijay
⇒ (1400 + x) : (1800 + 2x) = (5600 - 3200) : 3200
⇒ (1400 + x) /(1800 + 2x) = (5600-3200)/3200
⇒ (1400 + x) /(1800 + 2x) = 2400 / 3200
⇒ (1400 + x) /(1800 + 2x) = 3 / 4
⇒ 5600 + 4x = 5400 + 6x
⇒ 6x - 4x = 5600 - 5400
⇒ 2x = 200
∴ x= 100

∴ Initial investment of Ajay = 1400 + 100 = Tk.1500

Question: If A = {1, 2, 3} and B = Ø, what is the value of (A U B)?

Solution:
দেওয়া আছে,
A = {1, 2, 3}
এবং B = Ø
যেকোনো সেট A এবং ফাঁকা সেট (Ø) এর সংযোগ (union) হলো A 

∴ (A ∪ B) = {1, 2, 3} ∪ Ø
= A
= {1, 2, 3}

Question: If 26 August of a certain year was Tuesday, what day was 29 September of that year?

Solution:
From 26 August to 29 September,
= 6 + 29 days
= 35 days

The same weekday repeats after every 7 days (or multiples of 7).

So, the 36th day would be Tuesday again.
Therefore, the 35th day (29 September) was Monday.

Given,
9x + 3y + 12 = 0 and 18x + 6y + 26 = 0
a1/a2 = 9/18 = 1/2
b1/b2 = 3/6 = 1/2
c1/c2 = 12/26 = 6/13
Since, a1/a2 = b1/b2 ≠ c1/c2

So, the pairs of equations are parallel and the lines never intersect each other at any point, therefore there is no possible solution.

Question: An officer was appointed on maximum daily wages on contract money of Tk. 6720. But on being absent for some days, he was paid Tk. 5600. For how many days was he absent?

Solution:
Maximum daily wages of the officers = H.C.F of Tk. 6720 and Tk. 5600

H.C.F of 6720 & 5600 = 1120

Maximum daily wage = Tk. 1120
Total days appointed = 6720 ÷ 1120 = 6 days
​
​Absent days = 6 − 5 = 1 day

Question: Among 100 students, the average marks in English is 78. If the 60 girls scored an average of 84, determine the average score of the remaining boys.

Solution:
ধরি,
ছাত্রদের গড় নম্বর = x 

100 জন শিক্ষার্থীর মোট নম্বর = 100 × 78 = 7800 
এবং 60 জন ছাত্রীর মোট নম্বর = 60 × 84 = 5040

প্রশ্নমতে,
5040 + (100 - 60)x = 7800
⇒ 5040 + 40x = 7800
⇒ 40x= 7800 - 5040
⇒ 40x = 2760
⇒ ক = 2760/40 
⇒ ক = 69

∴ 40 জন ছাত্রের গড় নম্বর = 69

Let,
The speed is x km/hr.
Then,
Distance covered in 1 hr. 40 min
i.e., 1(2/3) hrs. = 5x/3 km.
Remaining Distance = {24 - (5x/3)}
∴ 5x/3 = 5/7{24 - (5x/3)
⇒ 5x/3 = 5/7{(72 - 5x)/3}
⇒ 7x = 72 - 5x
⇒ 12x = 72
⇒ x = 72/12
⇒ x = 6.
Hence, Speed = 6 km/hr
= {6 × (5/18)} m/s
= 5/3 m/s
= 1(2/3) m/s.

Question: If LOAD = DAOL, then ROAD = ?

Solution:
LOAD = DAOL এর L আর D এবং O আর A নিজেদের মধ্যে স্থান পরিবর্তন করেছে।

একইভাবে,
ROAD শব্দের R আর D এবং O আর A নিজেদের মধ্যে স্থান পরিবর্তন করেছে।

∴ ROAD = DAOR

Question: A team of 6 players is to be selected from 8 batsmen and 5 bowlers. How many ways can this be done if exactly 4 batsmen must be selected?

Solution:
এখানে, একটি দল গঠন করতে হলে,  4 জন ব্যাটসম্যান এবং 2 জন বোলার নির্বাচন করতে হবে।

8 জন ব্যাটসম্যান থেকে 4 জন ব্যাটসম্যান নির্বাচন করার উপায়:
⁸C₄ = 8!/{4! (8 - 4)!} 
= (8 × 7 × 6 × 5)/(4 × 3 × 2 × 1)
= 70 টি

5 জন বোলার থেকে 2 জন বোলার নির্বাচন করার উপায়:
⁵C₂ = 5!/{2! (5 - 2)!}
= (5 × 4)/(2 × 1)
= 10 টি

সুতরাং, মোট সম্ভাব্য উপায় = 70 × 10 = 700 টি।

অতএব, দল গঠনের মোট উপায় হলো 700 টি।

Question: P and Q are two alloys of gold and copper prepared by mixing metals in the ratio 7:2 and 7:11 respectively. If equal amounts of both alloys are melted to form a third alloy R, then the ratio of gold and copper in R will be - 
​
​Solution: 
Alloy P (Gold:Copper) = 7:2
Gold fraction in P = 7/9
Copper fraction in P = 2/9
​
​Alloy Q (Gold:Copper) = 7:11
Gold fraction in Q = 7/18
Copper fraction in Q = 11/18
​
Let, 1 kg of each alloy is mixed; 
Gold in Alloy R = 7/9 + 7/18 = 21/18
​
​Copper in Alloy R = 2/9 + 11/18 = 15/18 
​
​∴ The ratio of Gold:Copper in R = (21/18) : (15/18)
= 21 : 15
= 7 : 5

The word 'LEADING' has 7 different letters.
When the vowels EAI are always together, they can be supposed to form one letter.
Then, we have to arrange the letters LNDG (EAI).
Now, 5 (4 + 1) letters can be arranged in 5! = 120 ways.
The vowels (EAI) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6)
= 720.