Bank Math

Question: If 8 people make 48 chairs in 12 days by working 6 hours a day, then how long will it take 12 people working 8 hours a day to make 192 chairs?

Solution:
48 টি চেয়ার তৈরি করতে 8 জন লোক প্রতিদিন 6 ঘণ্টা কাজ করে = 12 দিনে
∴ 1 টি চেয়ার তৈরি করতে 1 জন লোক প্রতিদিন 1 ঘণ্টা কাজ করে = (8 × 6 × 12)/48 দিনে
∴ 192 টি চেয়ার তৈরি করতে 12 জন লোক প্রতিদিন 8 ঘণ্টা কাজ করে = (576 × 192)/(48 × 8 × 12)
= 24 দিনে

Question: Raju, Bappa and Samir divide Tk. 1,500 among them in such a way that if Tk. 50, Tk. 40 and Tk. 60 are removed from the sums that Raju, Bappa and Samir received respectively, then the shares will be in the ratio of 2 : 3 : 4. How much did Raju receive?

সমাধান:
ধরি, 
​টাকা সরিয়ে নেওয়ার পর রাজু, বাপ্পা এবং সমীরের প্রাপ্ত অংশের অনুপাত যথাক্রমে 2x, 3x এবং 4x।

তাহলে, টাকা সরিয়ে নেওয়ার আগে তাদের প্রাপ্ত অংশ ছিল:
রাজু = 2x + 50
বাপ্পা = 3x + 40
সামীর = 4x + 60

তাদের প্রাপ্ত মোট টাকা হলো Tk. 1,500।
​প্রশ্নমতে,
⇒ (2x + 50) + (3x + 40) + (4x + 60) = 1500
⇒ (2x + 3x + 4x) + (50 + 40 + 60) = 1500
⇒ 9x + 150 = 1500
⇒ 9x = 1500 - 150
⇒ 9x = 1350
⇒ x = 1350/9
∴ x = 150

রাজুর প্রাপ্ত টাকা = 2x + 50
= 2(150) + 50
= 300 + 50
= 350
সুতরাং, রাজু Tk. 350 পেয়েছিল।

Let, B alone filled the pipe by x hours.

Efficiency ratio of B and C =2 : 1
Time ratio of B and C = 1 : 2

Given,
(1/A + 1/B)- (1/A + 1/C) =7/60 - 1/12
⇒ 1/B - 1/C = 2/60 = 1/30
⇒ 1/x - 1/2x=1/30
⇒ 1/2x=1/30
⇒ 1/x=1/15
⇒ x = 15

B alone filled the pipe by 15 hours.

Question: A box contains 400 marbles, of which 40% are blue. You pick some marbles of which 25% are blue. Of the remaining marbles, 50% are blue marbles. How many marbles did you pick?

Solution:
Given that, 
Total marbles = 400
Blue marbles initially = 40% of 400 = (40/100) × 400 = 160
Non-blue marbles = 400 - 160 = 240

Let the number picked be x.
∴ Blue picked = 25% of x = 0.25x
and non-blue picked = 0.75x

And, 
Remaining marbles = 400 - x
∴ Remaining blue marbles = 160 - 0.25x

ATQ, Of the remaining marbles, 50% are blue then we get, 
160 - 0.25x = 0.5(400 - x)
⇒ 160 - 0.25x = 200 - 0.5x
⇒ 0.5x - 0.25x = 200 - 160
⇒ 0.25x = 40
⇒ x = 40/0.25
∴ x = 160

So the number of marbles you picked is 160.

Question: One tap (A) fills a reservoir four times as fast as another tap (B). If both taps running together can fill the reservoir in 20 minutes, then how long will the slower tap (B) alone take to fill the reservoir?

সমাধান:
ধরি,
ধীরগতির নল B একা চৌবাচ্চাটি পূর্ণ করতে সময় নেয় x মিনিট।
তাহলে, দ্রুতগতির নল A একা চৌবাচ্চাটি পূর্ণ করতে সময় নেবে x/4 মিনিট।

প্রশ্নমতে, তারা একত্রে 20 মিনিটে পূর্ণ করে। অর্থাৎ,
1/x + 1/(x/4) = 1/20
⇒ 1/x + 4/x = 1/20
⇒ (1 + 4)/x = 1/20
⇒ 5/x = 1/20
⇒ x = 5 × 20
⇒ x = 100 মিনিট
∴  x = 1 ঘণ্টা 40 মিনিট [ 60 মিনিট = 1 ঘণ্টা]

∴ ধীরগতির নলটি (B) একা চৌবাচ্চাটি পূর্ণ করতে 1 ঘন্টা 40 মিনিট সময় নেবে।

Question: The average of 5 consecutive numbers is n. What will be the average if the next two numbers are included?

Solution:
The average of 5 consecutive terms is n, implies that the 3rd term is n. Now as the next 2 terms are included implies that the new average for 7 terms would be the 4th term. So, the 4th term would be n + 1.

Example:
(1 + 2 + 3 + 4 + 5)/5
= 15/5
= 3

(1 + 2 + 3 + 4 + 5 + 6 + 7)/7
= 28/7
= 4

Question: If the difference between the circumference and diameter of a circle is 120 cm, then the diameter of the circle is -

Solution:
ধরি,
বৃত্তের ব্যাসার্ধ = r
বৃত্তের ব্যাস = 2r
বৃত্তের পরিধি = 2πr

প্রশ্নমতে,
2πr - 2r = 120
⇒ 2r(π - 1) = 120
⇒ r = (120/2){(22/7) - 1}
⇒ r = 60/(22 - 7)/7
⇒ r = (60 × 7)/15
∴ r = 28

∴ বৃত্তের ব্যাস = 2r = 2 × 28 = 56 সে.মি.

Question: In a simultaneous throw of a pair of dice, what is the probability of getting a total more than 8?

Solution: 
The total number of possible outcomes for the pair of dice is the product of the outcomes for each die is-
= 6 × 6 = 36

And,
The favorable outcomes are the combinations where the sum of the two dice is more than 8. These sums can be 9, 10, 11, or 12. The combinations for each sum are-
Sum of 9: (3, 6), (4, 5), (5, 4), (6, 3) - 4 outcomes
Sum of 10: (4, 6), (5, 5), (6, 4) - 3 outcomes
Sum of 11: (5, 6), (6, 5) - 2 outcomes
Sum of 12: (6, 6) - 1 outcome

∴ The total number of favorable outcomes is the sum of these outcomes = 4 + 3 + 2 + 1 = 10

∴ P(sum > 8) = Favorable Outcomes/Total Outcomes = 10/36 = 5/18

Question: Find the domain of f(x) = 1/x + 3

Solution:
দেওয়া আছে,
f(x) = 1/x + 3

আমরা জানি, 
একটি ভগ্নাংশের হর (denominator) শূন্য হতে পারবে না।
অর্থাৎ,
⇒ x + 3 ≠ 0
∴ x ≠ - 3

সুতরাং, f(x) এর ডোমেইন = R - {- 3}

4 6 9 6 14 6 .......
এখানে, ২য়, ৪র্থ, ৬ষ্ঠ, ৮ম ......... ইত্যাদি অবস্থানে নির্দিষ্ট সংখ্যা 6 বিদ্যমান
অবশিষ্ঠ সংখ্যাগুলোর ধারাঃ 4     9          14        19
পার্থক্যঃ                           5         5          5
সুতরাং নির্ণেয় সংখ্যাটি হচ্ছে 14 + 5 = 19.

Question: The sum of the squares of three numbers is 532 and the ratio of the first and the second as also of the second and the third is 3 : 2. The third number is -

Solution:
Given that,
First : Second = 3 : 2
Second : third = 3 : 2
= {3 × (2/3)} : {2 × (2/3)}
= 2 : (4/3)

∴ Ratio between the numbers = 3 : 2 : (4/3)
= 9 : 6 : 4
Let the numbers be 9x, 6x and 4x
Then,
⇒ (9x)² + (6x)² + (4x)² = 532
⇒ 81x² + 36x² + 16x² = 532
⇒ 133x² = 532
⇒ x² = 4
⇒ x = 2

So, third number = 4x = 4 × 2 = 8

Question: How many years will it take for an investment of Tk.10000 to earn Tk. 1200 in simple interest rate of 6%?
(Officer Cash 2022 অনুযায়ী)

Solution: 
Given that,
Principal, P = 10000
Simple Interest, SI = 1200
Rate of interest, r = 6%
Time, n = ?

We know,
n = I/Pr
= 1200/(10000 × 6%)
= (1200 × 100)/(10000 × 6)
= 2

So, it will take 2 years for the investment to earn Tk. 1200 at 6% simple interest.

Let the age of son = X years

∴Age of mother would be =2X

As per question 20 years ago;
10 (X -20) = 2X - 20
⇒ 10X - 200 = 2X - 20
⇒ 10X - 2X= - 20 + 200
⇒ 8X = 180
⇒ X= 180/8
= 22.5 years

∴Age of mother = 22.5 × 2 = 45 years.

Question: A and B are two alloys in which the ratios of gold and copper are 5 : 3 and 5 : 11 respectively. If these equal amounts of two alloys are melted and made into alloy C. What will be the ratio of gold and copper in alloy C?

Solution:
Ratio of Gold and Copper in Alloy A = 5 : 3
Ratio of Gold and Copper in Alloy B = 5 : 11

Amount of Gold in Alloy A = 5/8
Amount of Gold In Alloy B = 5/16

Amount of Copper in A = 3/8
Amount of Copper in B = 11/16

∴ Amount of Gold In C = Amount of gold in A + Amount of gold in B 
= (5/8) + (5/16)
= (10 + 5)/16
= 15/16

∴ Amount of Copper in C = Amount of Copper in A + Amount of Copper in B
= (3/8) + (11/16)
= (6 + 11)/16
= 17/16

∴ Ratio of Gold and Copper in C = (15/16) : (17/16)
= 15 : 17

Ratio of capitals = 7/2 : 4/3 : 6/5 = (7/2 × 30) : (4/3 × 30) : (6/5 × 30)
= 105 : 40 : 36.
Let the initial capitals of A, B and C be Tk. 105x, 40x, 36x respectively.
Then, ratio of profit = [105x × 4 + (150% of 105x) × 8] : (40x × 12 ) : (36x × 12)
1680 : 480 : 432
= 35 : 10 : 9.
∴ A's share = Tk. (2430 × 35/54) = Tk. 1575; B's share = Tk (2430 × 10/54) = Tk. 450
C's share = Tk. (2430 × 9/54) = Tk. 405.

Question: A two-digit number has 5 in its ten's digit. The sum of its digits is one-sixth of the number itself. What is the number?

Solution:
ধরি,
একক স্থানীয় অঙ্ক = x
দশক স্থানীয় অঙ্ক = 5

∴ সংখ্যাটি = 50 + x

প্রশ্নমতে,
5 + x = (50 + x)/6
⇒ 6(5 + x) = 50 + x
⇒ 30 + 6x = 50 + x
⇒ 6x - x = 50 - 30 
⇒ 5x = 20
∴ x = 4

∴ সংখ্যাটি = 50 + 4 = 54

Let, the cost price be = 100 and selling price = 115
Their ratio = sp : cp = 115 : 100 = 23 : 20

Question: The clock showed 10 : 30 in the mirror. What is the actual time?

Solution:
আমরা জানি,
প্রকৃতপক্ষে সময়
= 11 : 60 - আয়নায় দেখা সময়
= 11 : 60 - 10 : 30
= 1 : 30

Let
amount of water be 8x.
So, the amount of sugar is 3x.
According to question,
8x/(3x + 2) = 2/1
Solving this equation, we get, x = 2
Therefore, the amount of sugar in the original solution = 3 × 2 = 6 kg.

Question: The diagonal of a rectangle is √41 cm and its area is 20 sq. cm. The perimeter of the rectangle must be:

Solution:
Let the length and breadth of the rectangle be x cm and y cm.
We are given two conditions are,
Diagonal = √41 cm
By Pythagoras theorem:
√(x² + y²) = √41
⇒ x² + y² = 41 ........ (1)
And Area, xy = 20 cm² .......(2)

We know,
(x + y)² = x² + y² + 2xy
= 41 + 2 × 20
= 41 + 40
= 81
⇒ x + y = √81
∴ x + y = 9

∴ Perimeter = 2(x + y) = 2 × 9 = 18 cm

 So the perimeter of the rectangle is 18 cm.

Between 100 and 200 (and including 100) there are 21 numbers evenly divisible by 5. 
201 to 900 only net 20 numbers per century range that qualify. So, now there are 20×7 = 140 numbers evenly divisible by 5.

In last 100 (901 to 999) we only have 19 (the 20th one is 1,000 which is not a 3 digit number).

So, in total, we have 21 + 140 + 19 = 180

Question: Which value of y will satisfy the given inequality, 2(y - 3) ≥ 3y - 4 ?

Solution:
Given,
2(y - 3) ≥ 3y - 4
⇒ 2y - 6 ≥ 3y - 4
⇒ 2y - 3y ≥ - 4 + 6
⇒ - y ≥ 2
⇒ y ≤ - 2

Question: If 7^b = 343, then the value of 7^{(b - 2)} is:

Solution:
Given that,
7^b = 343
⇒ 7^b = 7³
∴ b = 3

Now,
7^{(b - 2)}
= 7^{(3 - 2)}
= 7¹
= 7 

Question: A 320 metre long train crosses a platform twice its length in 48 seconds. What is the speed of the train in km/hr?

Solution:
দেওয়া আছে,
ট্রেনটির দৈর্ঘ্য = 320 মিটার
প্ল্যাটফর্মটির দৈর্ঘ্য = 2 × 320 = 640 মিটার

অতিক্রান্ত মোট দূরত্ব = (ট্রেনের দৈর্ঘ্য + প্ল্যাটফর্মের দৈর্ঘ্য)
= (320 + 640) মিটার
= 960 মিটার

সময় লেগেছে = 48 সেকেন্ড

∴ ট্রেনটির গতিবেগ = দূরত্ব / সময়
= 960 / 48 মিটার/সেকেন্ড
= 20 মিটার/সেকেন্ড

এখন,
1 মিটার/সেকেন্ড = (1/1000)/(1/3600) কিমি/ঘন্টা
= 3.6 কিমি/ঘন্টা

1 মিটার/সেকেন্ড = 3.6 কিলোমিটার/ঘন্টা
∴ 20 মিটার/সেকেন্ড = 20 × 3.6 কিলোমিটার/ঘন্টা
= 72 কিলোমিটার/ঘন্টা

সুতরাং, ট্রেনটির গতিবেগ 72 কিমি/ঘন্টা।

Here,
we just know the selling price and the gain and loss incurred, on two cameras.
Therefore,
first calculate the cost price of both the cameras, because gain or loss is calculated on the cost price of the material.
When a shopkeeper earns profit, Cost Price = 100/(100 + Gain%) × S.P.
When shopkeeper incurs a loss, Cost Price = 100/(100 – Loss%) × S.P.

C.P. of camera A = 100/(100 +20) × 7500
= 100/120 × 7500
= Tk. 6250
C.P of camera B = 100/(100 – 5) × 8550
= 100/95 × 8550
= Tk. 9000

Total C.P. = Cost of camera A + Cost of camera B
Total C.P. = 6250 + 9000 = Tk. 15250
Total S.P. = 7500 + 8550 = Tk. 16050
Selling Price > Cost Price, hence man gains during this transaction.

Gain = S.P. – C.P.
= 16050 – 15250
= Tk. 800
Gain% = Gain/C.P. × 100
Gain% = 800/15250 × 100
= 5.24%

Semi perimeter, s = (3 + 5 + 6)/2
= 7 cm
∴ Area = √{s(s - a)(s - b)(s - c)}
= √{7 (7 - 3) (7 - 5) (7 - 6)} Sq cm
= √(7 × 4 × 2 × 1)Sq cm
= 2√14 Sq cm

Question: Find HCF of 84, 126, and 210.

Solution:
Factorize each number into prime factors:
84 = 2² × 3 × 7
126 = 2 × 3² × 7
210 = 2 × 3 × 5 × 7

HCF is the product of common prime factors with smallest powers:
Common primes: 2, 3, 7
Smallest powers: 2¹, 3¹, 7¹
HCF = 2 × 3 × 7 = 42

Say x is the height of the building.
a is a point 30 m away from the foot of the building.
Here, height is the perpendicular and distance between point a and foot of building is the base.
The angle of elevation formed is 30°.

Hence,
tan 30° = perpendicular/base = x/30
1/√3 = x/30
x = 30/√3

Question: A boat takes 8 hours to travel 32 km upstream (against the current). If it were traveling downstream (with the current), it would take only 4 hours to cover the same distance. What is the speed of the current?

Solution:
দেওয়া আছে,
স্রোতের প্রতিকূলে 32 কিমি যেতে সময় লাগে 8 ঘণ্টা।
∴ প্রতিকূলে নৌকার গতিবেগ = 32/8 = 4 কিমি/ঘণ্টা।

আবার, স্রোতের অনুকূলে 32 কিমি যেতে সময় লাগে 4 ঘণ্টা।
∴ অনুকূলে নৌকার গতিবেগ = 32/4 = 8 কিমি/ঘণ্টা।

আমরা জানি,
স্রোতের গতিবেগ = (অনুকূলে গতিবেগ - প্রতিকূলে গতিবেগ) / 2
= (8 - 4)/2 কিমি/ঘণ্টা
= 4/2 কিমি/ঘণ্টা
= 2 কিমি/ঘণ্টা

সুতরাং, স্রোতের গতিবেগ 2 কিমি/ঘণ্টা।

Question: If sec(3x - 40°) = cosec(50° - x), then the value of x is?

Solution:
sec(3x - 40°) = cosec(50° - x)
⇒ sec(3x - 40°) = cosec{90° - (40° + x)}
⇒ sec(3x - 40°) = sec(40° + x)
⇒ 3x - 40° = 40° + x
⇒ 2x = 80°

∴ x = 40°

Question: If LATER = 13579 and CHAIR = 20349, then CHEAT = ?

Solution:
Given that,
L    A   T   E    R
↓    ↓   ↓    ↓    ↓
1    3   5   7    9

and
C   H   A   I    R
↓    ↓   ↓    ↓    ↓
2   0    3   4    9

So
C   H   E    A   T
↓    ↓   ↓    ↓    ↓
2   0    7   3    5

SI =  10000 × 2 × 5/100 = 1000
CI = 10000(1 + 5/100)² – 10000 = 1025
∴ Difference =  1025 - 1000 =  25

Question: Two inlet pipes can fill a tank in 10 hours and 20 hours, respectively. An outlet pipe is attached to these two pipes, and thus, the tank was filled in 12 hours. In 90 hours, the outlet pipe alone can empty how many tanks?

সমাধান:
ধরি,
ছিদ্র নলটি (Outlet Pipe) একা ট্যাঙ্কটি খালি করতে P ঘন্টা সময় নেয়।

তিনটি নল একত্রে 1 ঘন্টায় পূর্ণ করে = 1/10 + 1/20 - 1/P অংশ।
প্রশ্নমতে, তিনটি নল একত্রে 12 ঘন্টায় পূর্ণ করে।
∴ 1/10 + 1/20 - 1/P = 1/12

১. ছিদ্র নলটির সময় (P) নির্ণয়:
⇒ 1/P = 1/10 + 1/20 - 1/12
হরগুলির (Denominator) ল.সা.গু. (LCM) হলো 60।
⇒ 1/P = (6 + 3 - 5)/60
⇒ 1/P = 4/60
⇒ 1/P = 1/15
⇒ P = 15 ঘন্টা।

90 ঘন্টায় যতগুলি ট্যাঙ্ক খালি করতে পারে = 90 / P
= 90/15
= 6 টি ট্যাঙ্ক।
∴ 90 ঘন্টায় ছিদ্র নলটি একা 6 টি ট্যাঙ্ক খালি করতে পারে।

Question: A square is inscribed in a circle of diameter 2a and another square is circumscribing circle. The difference between the areas of outer and inner squares is-

Solution:

দেওয়া আছে 
বৃত্তের ব্যাস = 2a
অভ্যন্তরীণ বর্গের কর্ণ = 2a 
অভ্যন্তরীণ বর্গের এক বাহু = x 

প্রশ্নমতে
√2x = 2a 
x = 2a/√2
x = √2a

অভ্যন্তরীণ বর্গের ক্ষেত্রফল = (√2a)²
= 2a²

অভ্যন্তরীণ বর্গের কর্ণ = বহিঃস্থ বর্গের এক বাহু = 2a
বহিঃস্থ বর্গের ক্ষেত্রফল = (2a)² = 4a² 

ক্ষেত্রফলের পার্থক্য = 4a² - 2a²
= 2a²

Let,
The edge of the third small cube be x cm
The volume of the cube = (edge)³
According to the question,
6³ + 8³ + x³ = 12³
⇒ 216 + 512 + x³ = 1728
⇒ x³ = 1728 - 728
= 1000
⇒ x = 1000^(1/3)
⇒ x = 10 cm.

Question: When the selling price is doubled, the profit becomes threefold. What is the percentage profit?

Solution:
ধরি,
ক্রয়মূল্য = 100 টাকা 
বিক্রয়মূল্য = (100 + x) টাকা 

∴ লাভ = (100 + x) - 100 টাকা = x টাকা

প্রশ্নমতে,
3x = 2(100 + x) - 100
⇒ 3x = 200 + 2x - 100
⇒ 3x = 100 + 2x
⇒ 3x - 2x = 100 
⇒ x = 100

∴ লাভ = 100 টাকা

এখন,
লাভের হার = (লাভ/ক্রয়মূল্য) × 100%
= (100/100) × 100%
= 100%

∴ লাভের শতকরা হার = 100%

Question: Four bells ring simultaneously at the start and then at intervals of 6 seconds, 12 seconds, 15 seconds, and 20 seconds respectively. How many times do they ring together in 2 hours?

Solution:
Given that,
Four bells ring simultaneously at starting and an interval of 6 sec, 12 sec, 15 sec and 20 sec respectively.

∴ LCM of (6, 12, 15, 20) = 60
All 4 bells ring together again after every 60 seconds

Now, In 2 Hours, they ring together = [(2 × 60 × 60)/60] times + 1 (at the starting)
= (120 + 1) times
= 121 times

∴ In 2 hours they ring together for 121 times

r = √(13² - 12²)
= 5

So, the volume is V = 1/3∏r²h
= 1/3Π × 5² × 4
= 33.33Π cm³

Question: If ⁿC₁₀ = ⁿC₈ , what is the value of ⁿC₂ = ?

Solution:
আমরা জানি,
যদি ⁿC_a = ⁿC_b হয়, তবে হয় a = b অথবা a + b = n হবে।

এখানে,
ⁿC₁₀ = ⁿC₈
⇒ 10 + 8 = n
⇒ n = 18

∴ ⁿC₂ = ¹⁸C₂
= 18!/(2!(18 - 2)!)
= 18!/(2! × 16!)
= (18 × 17 × 16!)/(2 × 1 × 16!)
= (18 × 17)/2
= 9 × 17
= 153

Question: If dividing P(x) = 4x³ - 7x² + bx - 5 by (x - 2) results in the remainder 13, then find the value of b.

Solution:
According to the Remainder Theorem, if a polynomial P(x) is divided by (x - c), then the remainder = P(c).
Here divisor is (x - 2).
So remainder = P(2).

Now,
P(2) = 4(2)³ - 7(2)² + b(2) - 5

= 4 × 8 - 7 × 4 + 2b - 5

= 32 - 28 + 2b - 5

= 32 + 2b - 33
= 2b - 1

According to the question, the remainder is 10.
So, 2b - 1 = 13

⇒ 2b = 13 + 1
⇒ 2b = 14
⇒ b = 7

Question: The population of a town is 20,000, and it increases by 5% each year. What will the population be after 3 years?

Solution:
We can use the compound interest formula for population growth:

Population after n years = P × [1 + (r/100)]ⁿ

Here,
P = 20,000, r = 5%, n = 3

∴ Population after 3 years = 20,000 × [1 + (5/100)]³
= 20,000 × (105/100)³
= 20,000 × 1.157625
= 23,152.5

∴ The population of the town after 3 years will be approximately 23,153.

As, Mr. x is always chosen, then to form a committee of 5 members from the pool of 8, we now have to choose 4 members from 7 people

So, the committee can be formed in ⁷c₄ = 35 ways

Question: A committee of 5 members is to be selected from 7 men and 4 women. In how many ways can this be done if exactly 3 men must be selected?

Solution:
এখানে, একটি কমিটি গঠন করতে হলে 3 জন পুরুষ এবং (5 - 3) = 2 জন মহিলা নির্বাচন করতে হবে।

7 জন পুরুষ থেকে 3 জন পুরুষ নির্বাচন করার উপায়:
⁷C₃ = 7!/{3! (7 - 3)!}
= (7 × 6 × 5)/(3 × 2 × 1)
= 35 টি

4 জন মহিলা থেকে 2 জন মহিলা নির্বাচন করার উপায়:
⁴C₂ = 4!/{2! (4 - 2)!}
= (4 × 3)/(2 × 1)
= 6 টি
সুতরাং, মোট সম্ভাব্য উপায় = 35 × 6 = 210 টি।

অতএব, কমিটি গঠনের মোট উপায় হলো 210 টি।