Bank Math

Question : A starts a business with Tk 4,500. After 4 months, B joins as a partner by investing some amount. At the end of one year from the start, the profits are shared in the ratio A : B = 3 : 4. Find B’s capital investment.

Solution:
Let,
B's capital be TK X
Then,
(4500 × 12)/8X = 3/4
⇒ 24X = (4500 × 12 × 4)
⇒ 24X = 216,000
⇒ X = 216,000/24
∴ X = 9000

Question: Two unbiased coins are tossed. What is the probability of getting at most one head?

Solution:
Total cases = {HH, HT, TH, TT} = 4
Favorable cases = {TT, HT, TH} = 3

∴ Required Probability = 3/4

Question: A book seller allowed 10% discount on printed price. He gets 30% commission from publisher. His profit in percent will be?

Solution: 
let, the marked price of book is 100 taka 
buying price 100 × 0.7 taka 
= 70 taka 

selling price of book seller = 100 × 0.9 taka 
= 90 taka 

Profit in percent = {(90 - 70)/70} × 100% 
= (200/7)% 

Question: A trader mixes 26 kg of rice at Tk 20 per kg with 30 kg of rice of other variety at Tk. 36 per kg and sells the mixture at Tk. 30 per kg. His profit percent is:

Solution: 
C.P. of 56 kg rice = (26 × 20 + 30 × 36)
= (520 + 1080)
=1600 taka

S.P. of 56 kg rice = (56 × 30)
= 1680 taka

∴ Profit = 1680 - 1600
= 80 taka

 ∴ Profit percentage = (80/1600) × 100% = 5%

Question: The average age of a group of 12 students is 18 years. When 3 more students join the group, the overall average age increases by 1 year. What is the average age of the three new students?

Solution:
Total age of 12 students = 12 × 18 = 216 years
After 3 new students joined, total students = 12 + 3 = 15
New average = 18 + 1 = 19 years
Total age of 15 students = 15 × 19 = 285 years
Total age of 3 new students = 285 - 216 = 69 years
Average age of 3 new students = 69 ÷ 3 = 23 years

∴ The average age of the three new students is 23 years.

Question: Find the value of cos(5π/6).

Solution:
cos(5π/6)
= cos(π - π/6) [∵ (π - θ) দ্বিতীয় চতুর্ভাগে পড়ে এবং দ্বিতীয় চতুর্ভাগে cos ঋণাত্মক, তাই cos(π - θ) = -cos θ]
= - cos(π/6)
= - cos(30°)
= - √3/2

Let man's rate upstream be x km/hr
Then, his rate downstream = 2x km/hr
∴ (speed in still water) : (Speed of stream)
(2x + x)/2 : (2x - x)/2
3x/2 : x/2
3 : 1

Question: If the difference between the circumference and diameter of a circle is 120 cm, then the radius 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(π - 1)}
⇒ r = 60/{(22/7) - 1}
⇒ r = 60/{(22 - 7)/7}
⇒ r = 60/(15/7)
⇒ r = 60 × (7/15)
⇒ r = 4 × 7
⇒ r = 28

∴ বৃত্তের ব্যাসার্ধ = 28 সে.মি.

Question: The sum of the ages of a mother and daughter is 50 years. Five years ago, the mother was four times as old as her daughter. What is the present age of the daughter?

Solution:
ধরি, মেয়ের বর্তমান বয়স = x বছর
তাহলে, মাতার বর্তমান বয়স = (50 - x) বছর

পাঁচ বছর আগে,
মেয়ের বয়স ছিল = (x - 5) বছর
মাতার বয়স ছিল = (50 - x) - 5 = (45 - x) বছর

প্রশ্নমতে,
45 - x = 4(x - 5)
⇒ 45 - x = 4x - 20
⇒ 45 + 20 = 4x + x
⇒ 65 = 5x
⇒ x = 65/5
⇒ x = 13

সুতরাং, মেয়ের বর্তমান বয়স হলো 13 বছর।

'DARKER' has 6 letters.

Thus, we can arrange 6 letters in 6! ways.
But R gets repeated. There are 2R's. So divide by 2!

∴ Total ways = 6!/2! = 360
Vowels not together = Total ways - Vowels together

Consider the 2 vowels (A and E) as one group.
We have 4 letters and 1 group = 5
We can arrange them in 5! Ways.

But again here R comes twice. So we will have 5!/2!
Also, the 2 vowels can be arranged in 2! Ways.
SO the number of ways with vowels together = 2! × (5!/2!) = 120

∴ Number of ways with vowels not together = 360 - 120 = 240.

Total attendance for first 3 days = 325×3 = 975
Total attendance for first 4 days = 320×4 = 1280
∴ Present on the 4th day = 1280 - 975 = 305

(2√27 – √75 + √12)
= 2√(3².3) - √(3 × 5²) + √(3 × 2²)
= 6√3 – 5√3 + 2√3
= 3√3

• 2 men and 3 boys can do a piece of work in 10 days.
Thus, 20 men and 30 boys can do a piece of work in 1 day......(i)

• 3 men and 2 boys can do the same work in 8 days.
Thus, 24 men and 16 boys can do the same work in 1 day....(ii)

• Equating (i) and (ii) we get -
o 20 men + 30 boys = 24 men + 16 boys
o 4 men = 14 boys
o 2 men = 7 boys

• Substituting this in equation (i) we get

10 boys can do a piece of work in 10 days.
But we need to find out in how many days 2 men and 1 boy can do the work, which is equivalent to 8 boys.

8 boys can do the same work in (10 × 10/8) = 12.5 days.

Question: A woman purchased 500 shares of the face value of Tk. 100 each from the market at Tk. 160 per share. If a dividend of 20% is declared, find her earning percent on the investment.

Solution:
Given,
Price of 1 share (Market Value) = Tk. 160
∴ Price of 500 shares (Total Investment) = Tk. (160 × 500) = Tk. 80,000

Dividend per share = 20% of 100 = Tk. 20
∴ Total dividend income = 500 × 20 = Tk. 10,000

∴ Earning % = (Total Dividend/Total Investment) × 100
= (10,000/80,000) × 100
= (1/8) × 100
= 12.5%

Let,
Jashim's present age be x
Jashim's age before 5 years = (x - 5)
Jashim's age after 10 years = (x + 10)
We are given that,
Jashim's age after 10 years (x + 10) is 5 times his age 5 years back (x – 5)

Therefore,
(x + 10) = 5 (x – 5)
⇒ x + 10 = 5x – 25
⇒ 4x = 35
⇒ x = 8.75 years.

Question: In a class if 5 students are seated in each bench, 5 benches remain vacant. But if 4 students are seated on each bench, 8 students are to remain standing. What is the number of students in that class?

Solution:
ধরি,
বেঞ্চ সংখ্যা x টি

একটি শ্রেণির প্রতি বেঞ্চে 5 জন করে ছাত্র বসলে 5 টি বেঞ্চ খালি থাকে।
∴ ছাত্রসংখ্যা= (x - 5) × 5 জন

প্রতি বেঞ্চে ৩ জন করে ছাত্র বসালে ৬ জন ছাত্রকে দাঁড়িয়ে থাকতে হয়।
∴ ছাত্রসংখ্যা = 4x + 8 জন

প্রশ্নমতে,
(x - 5) × 5 = 4x + 8
⇒ 5x - 25 = 4x + 8
⇒ 5x - 4x = 8 + 25 
∴ x = 33

∴ ছাত্রসংখ্যা = (x - 5) × 5 জন
= (33 - 5) × 5 জন 
= 28 × 5 জন 
= 140 জন 

∴ ঐ ক্লাসে ছাত্র সংখ্যা 140 জন। 

After 25 days, 35 men complete the work in 12 days.
Thus, 35 men can finish the remaining work in 12 days.
∴ 30 men can do it in (12 × 35)/30
= 14 days.
which is 1 day behind.

The present age of the son is P, he was born P years ago.
The age of the man was: (48 - P).

His age when the son was born should be equal to 2/3 of P.
(48 - P) = (2/3)P
5P = 144
⇒ P = 28.8 years.

Simple interest for 3 years
= 12005 - 9800
= 2205
Simple interest for 5 years
= (2205/3) × 5
= 3675
Some of money = 9800 - 3675
= 6125
R = (100 × 2205)/(6125 × 3)
= 12%

Question: Two cards are drawn at random and without replacement from a standard deck of 52 cards. What is the probability that both cards are face cards?

Solution:
Total card = 52
Total face card = 3 × 4 = 12

Total ways to choose 2 cards from 52 = ⁵²C₂ = (52 × 51)/2 = 1326

Total ways to choose 2 face cards from 12 = ¹²C₂ = (12 × 11)/2 = 66

∴ So, the probability that both cards are face cards = 66/1326
= (2 × 3 × 11)/(2 × 3 × 221)
= 11/221

Question: If x² + y² = 50 and xy = 21, what is the value of (x - y)²?
 
Solution:
We are given:
x² + y² = 50
xy = 21

Use the identity:
(x - y)² = x² + y² - 2xy

Substitute the values:
⇒ (x - y)² = x² + y² - 2xy
⇒ (x - y)² = 50 - 2 × 21
⇒ (x - y)² = 50 - 42
∴ (x - y)² = 8

Question: A cylindrical tank with a radius of 7 m and height of 2 m is filled with water. If the water is poured into a rectangular tank with a base measuring 7 m × 7 m, what will be the height of water in the rectangular tank?

Solution: 
Volume of the cylinder = π(7)²2
= π × 49 × 2
= 98π m³

Volume of the rectangle = 7 × 7 × h (Assuming, height of the rectangle is 'h')
= 49h m³ 

So, 49h = 98π
⇒ h = (98/49)π
∴ h = 2π m 

Salary Increased = 8100-7200 = 900
Percent increased = (900/7200) × 100 = 12.5%

Question: A student scores 55% marks in 8 papers, each carrying 100 marks. He obtains 15% of his total obtained marks in the Bangla paper. How many marks did he score in Bangla?

Solution:
৮ টি বিষয়ে মোট নম্বর = 8 × 100 = 800
সে নম্বর পেয়েছে = 800 এর 55%
= 800 এর 55/100
= 440

∴ সে বাংলায় নম্বর পেয়েছে = 440 এর 15%
= 440 এর 15/100
= 66

Question: Three numbers are in the ratio 1 : 2 : 3, and the sum of their cubes is 7776. The smallest number will be -

Solution:
Given,
The numbers be in the ratio 1 : 2 : 3
So, let:
Smallest number = x
Middle number = 2x
Largest number = 3x

ATQ,
x³ + 8x³ + 27x³ = 7776
⇒ 36x³ = 7776
⇒ x³ = 7776/36 = 216
∴ x = 6

So the smallest number is 6

Question: If x² - √7x + 1 = 0, then the value of x² + x^{- 2} = ?

Solution:
দেয়া আছে,
x² - √7x + 1 = 0
⇒ x² + 1 = √7x
⇒ x + (1/x) = √7 [উভয়পক্ষকে x দ্বারা ভাগ করে]

প্রদত্ত রাশি = x² + x^{- 2}
= x² + (1/x²)
= (x + 1/x)² - 2 . x . (1/x)
= (√7)² - 2
= 7 - 2
= 5

∴ নির্ণেয় মান হলো 5

Question: In how many years Tk. 1000 will become Tk. 1331 at compound interest rate of 10% per annum?

Solution:
Given,
P = Tk. 1000
C = Tk. 1331
r = 10% p.a

We know,
P(1 + r)ⁿ = C
⇒ 1000(1 + 10%)ⁿ = 1331
⇒ (1 + 1/10)ⁿ = 1331/1000
⇒ (11/10)ⁿ = (11/10)³
∴ n = 3

∴ required time = 3 years

Question: The slope of the line 3x + y = 5 is not the same as the slope of which one of the following lines?

Solution: 
প্রথমে, প্রদত্ত রেখাটির ঢাল নির্ণয় করতে হবে। রেখাটির সমীকরণকে y =mx + c তে রূপান্তর করতে হবে। এখানে 'm' হলো ঢাল (Slope)।

 3x + y = 5 
⇒ y = - 3x + 5

∴ এই রেখাটির ঢাল (m) হলো - 3

এবার, প্রদত্ত বিকল্পগুলোর প্রত্যেকটির ঢাল নির্ণয় করি:
ক) 3x + y = 2
⇒ y = - 3x + 2
∴ ঢাল -3

খ) x + (y/3) = 4
⇒ y/3 = - x + 4
⇒ y = - 3x + 12
∴ ঢাল - 3.

গ) y = - 3x + 1
∴ ঢাল - 3.

ঘ) x + 3y = 6
⇒ 3y = - x + 6
⇒ y = - 1/3x + 2
∴ ঢাল - 1/3.

∴ অপশন (ঘ) এর রেখার ঢাল মূল রেখার ঢাল থেকে ভিন্ন।

Question: In what ratio must a mixture of 30% alcohol strength be mixed with that of 50% alcohol strength so as to get a mixture of 45% alcohol strength?

Solution:
মনে করি, প্রথম প্রকারের x একক এবং দ্বিতীয় প্রকারের y একক মিশ্রিত করতে হবে।

প্রশ্নমতে,
প্রথম প্রকারের অ্যালকোহলের পরিমাণ = 30% of x = 0.30x
দ্বিতীয় প্রকারের অ্যালকোহলের পরিমাণ = 50% of y = 0.50y

মিশ্রণের মোট পরিমাণ = (x + y)
মিশ্রণে অ্যালকোহলের মোট পরিমাণ = 45% of (x + y) = 0.45(x + y)

শর্তমতে,
0.30x + 0.50y = 0.45(x + y)
⇒ 30x + 50y = 45(x + y)  [উভয় পক্ষকে 100 দ্বারা গুণ করে]
⇒ 30x + 50y = 45x + 45y
⇒ 50y - 45y = 45x - 30x
⇒ 5y = 15x
⇒ x/y = 5/15
⇒ x/y = 1/3
∴ x : y = 1 : 3

∴ The ratio is 1 : 3

Question: A cube has a base with a perimeter of 24 cm. What is its volume?

Solution:
দেওয়া আছে,
ঘনকের ভূমির পরিসীমা = 24 cm
আমরা জানি,
ঘনকের ভূমি একটি বর্গক্ষেত্র।

বর্গক্ষেত্রের পরিসীমা = 4 × বাহুর দৈর্ঘ্য

ধরি, ঘনকের প্রতিটি বাহুর দৈর্ঘ্য = a

প্রশ্নমতে,
4a = 24
⇒ a = 24/4
⇒ a = 6 cm

এখন,
ঘনকের আয়তন, V = a³ ঘন একক
= (6)³
= 216 cm³

সুতরাং, ঘনকটির আয়তন হল 216 cm³।

Question: ΔABC is a right-angled isosceles triangle, and ∠B is the right angle in the triangle. If AC measures 10√2, then which one of the following would equal the lengths of AB and BC, respectively?

Solution:
যেহেতু ABC একটি সমকোণী সমদ্বিবাহু ত্রিভুজ এবং ∠B হলো সমকোণ, তাই সমকোণের সাথে সংযুক্ত বাহু দুটি অর্থাৎ AB এবং BC এর দৈর্ঘ্য সমান হবে।

 পিথাগোরাসের উপপাদ্য অনুসারে,
AB² + BC² = AC²
⇒ BC² + BC² = (10√2)² [এখানে, AB = BC এবং AC = 10√2]
⇒ 2BC² = 10² × 2
⇒ BC² = 10² 
⇒ BC = 10

সুতরাং, AB এবং BC এর দৈর্ঘ্য যথাক্রমে 10 এবং 10।

All possible outcome = {HHH, HHT, HTT, HTH, THH, TTH, THT, TTT} =  8
It will be head every time, this occurs 1 time
∴ Probability =  1/8

Speed of the first train = 48 km/hr.
Let the length of the first train = 2x metre.

Speed of the second train = 42 km/hr.
Let the length of the second train = x metre.

Distance = (2x + x) = 3x metre.
Time = 12 seconds

Relative speed = 48 + 42 = 90 km/hr.
= 90 × (5/18) = 25 m/s.
3x = 25 × 12
⇒ x = 100 metre.

Length of the first train = 200
Time is taken to cross the platform = 45 seconds.
Speed of first train = 48 km/hr
= 48 × (5/18)
= 40/3 m/s.

Let the length of the platform = y metre.
Distance = 200 + y metre.
⇒ 200 + y = 45 × (40/3)
⇒ 200 + y = 600
⇒ y = 400 metre.

Total number of students = (26 + 7) - 1 = 32

According to the question,
Cost Price : Marked Price
(100 - Discount) : (100 + Profit)
100 - 10 : 100 + 12
90 : 112
45 : 56

Question: What is the angle between the hour and minute hands of a clock when it is 20 minutes past 8?

Solution:
20 minutes past 8 অর্থাৎ, 8 টা 20 মিনিট।
= 8 + (20/60) ঘন্টা
= 8 + (1/3)
= 25/3 ঘন্টা

আমরা জানি,
ঘণ্টার কাঁটা 12 ঘণ্টায় 360° ঘোরে।
∴ 1 ঘণ্টায় ঘোরে = 360°/12 = 30°
∴ 25/3 ঘন্টায় ঘোরে = (30° × 25)/3 = 250°

আবার,
মিনিটের কাঁটা 60 মিনিটে 360° ঘোরে।
∴ 1 মিনিটে ঘোরে = 360°/60 = 6°
∴ 20 মিনিটে ঘোরে = 20 × 6° = 120°

∴ ঘড়ির কাঁটা দুটির মধ্যবর্তী কোণ = | 250° - 120° | = 130°