Bank Math

Total area to be painted = 25×12 +2(10×12 + 10×25) = 1040 sqr.ft

A paints = 200/5 = 40 sqr.ft per day
B paints = 250/2 = 125 sqr.ft per day

A + B = 40 + 125 = 165 sqr.ft
Number of days = 1040/165 = 6(10/33)

Question: The salaries of A, B and C are in the ratio 1 : 3 : 4. If the salaries are increased by 5%, 10% and 10% respectively, then the increased salaries will be in the ratio -

Solution:
Given that
Salary has  in 1 : 3 : 4 ratio
Let
A's Salary = Tk. 100
B's Salary = Tk. 300
C's Salary = Tk. 400

Now,
5% increase in A's Salary,
A's new Salary = (100 + 5% of 100) = Tk. 105

B's Salary increases by 10%, Then,
B's new Salary = (300 + 10% of 300) = Tk. 330

C's Salary increases by 10%,
C's new Salary = (400 + 10% of 400) = Tk. 440

Then, ratio of increased Salary,
A : B : C = 105 : 330 : 440 = 21 : 66 : 88

Question: A, B, C enter into a partnership investing Tk. 40,000, Tk. 50,000 and Tk. 60,000 respectively. Total profit of A, B, C is Tk. 45,000. So B's profit is-

Solution:
A : B : C = 40000 : 50000 : 60000
A : B : C = 4 : 5 : 6

∴ Sum of the ratio = 4 + 5 + 6 = 15

∴ B's profit = 45000 × (5/15)
= 45000 × (1/3)
= 15,000

∴ B's profit is Tk. 15,000

Question: The values of k for equation 3x² - 6x + k = 0 to have real roots is —

Solution:
এখানে, 3x² - 5x + k = 0 সমীকরণকে ax² + bx + c = 0 সমীকরণের সাথে তুলনা করলে করি।

আমরা জানি,
বাস্তবমূলের ক্ষেত্রে,
b² - 4ac ≥ 0
⇒ (- 6)² - 4 × 3 × k ≥ 0
⇒ 36 - 12k ≥ 0
⇒ 12k ≤ 36
⇒ k ≤ 3

Question: The present ages of three cousins are in the ratio of 4 : 5 : 6. Five years ago, their total age was 60 years. In three years, what will be the age of the youngest cousin?

Solution: Present age ratio of three cousins is 4 : 5 : 6
Let, their age is 4X, 5X and 6X respectively.

ATQ,
4X - 5 + 5X - 5 + 6X - 5 = 60
Or, 15X - 15 = 60
Or, 15X = 75
Or, X = 5

∴ The present age of the youngest cousin is = 4X = 4 × 5 = 20 years.
In three years, his age will be = (20 + 3) = 23 years.

Here there are 2 cases
Case 1: accident happens at 50 km
Case 2: accident happens at 60 km
Difference between two cases is only for the 10 kms between 50 and 60.
Time difference of 10 minutes is only due to these 10 kms.
In case 1, 10 kms between 50 and 60 is covered at (3/4)^th Speed.
In case 2, 10 kms between 50 and 60 is covered at usual Speed.
So the usual Time 't' taken to cover 10 kms, can be found out as below.
(4/3)t – t = 10 mins [s = vt]
⇒ t = 30 mins, d = 10 kms
so usual
Speed = Distance/Time
= 10/30min
= 10/(1/2) km/hr.
= 20 km/hr.

Question: Mohan lent some amount of money at 8% simple interest and an equal amount of money at 10% simple interest each for two years. If his total interest was TK. 720, what amount was lent in each case ?

Solution:
Let the amount invested = TK. P
According to the questions,
⇔ (P × 8 × 2)/100 + (P × 10 × 2)/100 = 720
⇔ (16P + 20P)/100 = 720
⇔ 36P = 72000
⇔ P = 2000

Question: In a group of five men, no two men have the same age. The oldest man is 50 years old, and the youngest is 30 years old. If X is the average age of the men in the group, which of the following best indicates all and only possible values of X? (All ages are in whole numbers)

Solution: 
Here, 
Oldest man = 50 years
Youngest man = 30 years
Average age = X

Since no two people are the same age, the minimum average = (30 + 31 + 32 + 33 + 50)/5
= 176/5
= 35.2 
and the maximum average = (30 + 47 + 48 + 49 + 50)/5
= 224/5
= 44.8

That is, the value of X is greater than 35 and less than 45.
i.e. 35 < X < 45 

Let the numbers be 2x, 5x
ATQ,
(2x + 16) / (5x + 16) = 1/2
Or, 4x + 32 = 5x + 16
Or, x = 16
∴ The numbers: 2x = 2.16 = 32
5x = 5.16 = 80

Question: A boat covers 143 km upstream in 13 hours and the same distance downstream in 11 hours. What is the speed (in km/h) of the boat in still (without stream) water?

Solution:
Let, the speed of the boat in still water = x km/h
and the speed of the stream = y km/h

According to the question,
143/(x - y) = 13
⇒ 13(x - y) = 143
⇒ x - y = 11 ............(1)

Again,
143/(x + y) = 11
⇒ 11(x + y) = 143
⇒ x + y = 13 ............(2)

Adding (1) and (2):
x - y + x + y = 11 + 13
2x = 24
⇒ x = 12 km/h

We know,
Time taken by a train of length l metres to pass a stationary object of length b metres is
the time taken by the train to cover (l + b) metres.
Given that,
Time taken by a train of length 300 meters to pass a wall of length 600 meters is
the time taken by the train to covers (300+600) meters.
That is, 81 seconds.
We know that,
distance/time = speed
Therefore, required speed = (300+600)/81 m/sec
= 900/81 m/sec
Converting the above to km/hr
= (900/81) x (18/5) km/hr [m/s to km/hr conversion: a m/s = a x 18/5 km/hr.]
= 40 km/hr
Hence the answer is 40 km/hr.

Question: The diagonal of a rectangular field is 15 metres and the difference between its length its length and width is 3 metres. The area of the rectangular field is-

Solution:
Let l and b be the length and breadth of the rectangle respectively.
Then,
⇒ √(l² + b²) = 15
⇒ (l² + b²) = (15)2
⇒ l² + b² = 225

And,
∴  l - b = 3
⇒ (l - b)² = 9
⇒ l² + b² - 2lb = 9
⇒ 225 - 2lb = 9
⇒ 2lb = 216
∴ lb = 108

Hence, area of the field = lb = 108m²

Question: A man walks a at a rate of 10 mph. After every ten miles, he rests for 6 minutes. How much time does he take to walk 50 miles?

Solution:
The man needs time = 50/10 = 5 hours
= (5 × 60)
= 300 minutes

He will also rest 4 times after 10, 20, 30 and 40 miles
Total resting time = 4 × 6 = 24 minutes.

Total time = 300 + 24 = 324 minutes

Question: Find the simple interest on BDT 8,000 at 6% per annum for 9 months.

Solution: 
Given, P = 8000 Taka
n = 9 months = 0.75 years
r = 6% 

Simple Interest, I = Pnr
= 8000 × 0.75 × 6/100
= 80 × 4.5
= 360 Taka

Question: Average of ten positive numbers is x. If each number increases by 10%, then x-

Solution:
Let the ten positive numbers be n₁, n₂,..., n₁₀
Then, n₁ + n₂ + n₃ + ........... + n₁₀ = 10x ............(1)

According to the question,
each number is increased by 10%
10% = 10/100 = 0.1
Suppose the average of the ten numbers = X₁

Therefore, X₁ =( 1.1n₁ + 1.1n₂ + 1.1n₃ + ........... + 1.1n₁₀)/10
⇒ X₁ = 1.1(n₁ + n₂ + n₃ + ........... + n₁₀)/10
⇒ X₁ = 1.1X [We get from equation (1)]
⇒ X₁/ X = 1.1
⇒ X₁/X = (1.1 x 100)%
⇒ X₁/X =  110%

Total percentage gains = (110 - 100)% = 10%

∴ The new average increases by 10%.

Question: What is the absolute value (modulus) of x if - 8 < x < 2?

Solution:
Given that, 
- 8 < x < 2
⇒ - 8 + 3 < x + 3 < 2 + 3
⇒ - 5 < x + 3 < 5
∴ |x + 3| < 5

Question: A shopkeeper offers two successive discounts of 10% and 20% on an item marked at Tk. 1,200. What is the final selling price?

Solution: 
New Price after First Discount : 
1200 - 10% of 1200
= 1200 - 120
= 1080 Taka

Final Selling Price after Second Discount:
1080 - 20% of 1080
= 1080 - 216
= 864

So, Final Selling Price = Tk. 864

Question: What should be the value of "P" so that the expression (9 − 12x + Px²) becomes a perfect square?

Solution:
(9 − 12x + Px²)
= (3)² − 2 × 3 × 2x + (2x)² + Px² - (2x)²
= (3 - 2x)² + Px² - 4x²

∴ the expression becomes a perfect square if,
Px² - 4x² = 0
⇒ Px² = 4x²
∴ P = 4

Question: A train 300 m long is running at a speed of 90 km/hr. If it passes through a tunnel in 40 seconds, then the length of the tunnel is-

Solution:
ট্রেনের গতিবেগ = 90 কিমি/ঘন্টা
= (90 × 5/18) মিটার/সেকেন্ড
= 25 মিটার/সেকেন্ড

একটি সুড়ঙ্গ অতিক্রম করার সময় ট্রেনটি তার নিজের দৈর্ঘ্যের সাথে সুড়ঙ্গের দৈর্ঘ্যও অতিক্রম করে।
সুতরাং, মোট অতিক্রান্ত দূরত্ব = ট্রেনের গতিবেগ × সময়
= 25 মিটার/সেকেন্ড × 40 সেকেন্ড
= 1000 মিটার

ধরা যাক, সুড়ঙ্গটির দৈর্ঘ্য = L মিটার
মোট অতিক্রান্ত দূরত্ব = ট্রেনের দৈর্ঘ্য + সুড়ঙ্গের দৈর্ঘ্য
⇒ 1000 = 300 + L
⇒ L = 1000 - 300
⇒ L = 700 মিটার

সুতরাং, সুড়ঙ্গটির দৈর্ঘ্য হলো 700 মিটার।

Net height ascended in 2 min = (6 - 3) m = 3 m.
Net height ascended in 36 min = (3/2 × 36) = 54m.
In the 37th min,
the monkey ascends 6m and reaches the top.
Hence,
Total time taken = 37 minutes.

Number of rotation in one hour = 10 × 60 = 600
So, Distance moved = (600 × 20) = 12000 m

Question: A mother was 27 years old when her daughter was born. Currently, the mother's age is 9 years more than three times her daughter's age. How old will the daughter be in 6 years?

Solution:
Let the daughter's current age be x years.
∴ Mother's current age = 3x + 9 years.

According to the question,
3x + 9 - x = 27
⇒ 2x = 27 - 9
⇒ 2x = 18
∴ x  = 9

∴ Daughter's current age 9 years.
∴ Daughter's age after 6 years = 9 + 6 = 15 years.

Question: 12 workers can dig a 36-meter long trench working 8 hours per day. How many extra workers are required to dig a 45-meter long trench working 5 hours per day?

Solution:
8 ঘণ্টা কাজ করে 36 মিটার পরিখা খনন করে 12 জন শ্রমিক।
1 ঘণ্টা কাজ করে 1 মিটার পরিখা খনন করতে প্রয়োজন = (12 × 8)/36 জন শ্রমিক
5 ঘণ্টা কাজ করে 45 মিটার পরিখা খনন করতে প্রয়োজন = (12 × 8 × 45)/(36 × 5) জন শ্রমিক
= (12 × 8 × 45)/180 জন শ্রমিক
= 4320/180 জন শ্রমিক
= 24 জন শ্রমিক

∴ অতিরিক্ত শ্রমিকের সংখ্যা = 24 - 12 = 12 জন

Question: If principal M becomes N in 2 years when interest R% is compounded half-yearly. And if the same principal M becomes N in 2 years when interest S% is compound annually, then which of the following is true?

Solution:
ধরি, আসল = M এবং সবৃদ্ধি মূলধন = N
সময় n = 2 বছর

অর্ধবার্ষিক চক্রবৃদ্ধির ক্ষেত্রে (মুনাফার হার R%):
N = M(1 + (R/2)/100)^{2 × 2}
= M(1 + R/200)⁴

বার্ষিক চক্রবৃদ্ধির ক্ষেত্রে (মুনাফার হার S%):
N = M(1 + S/100)²

যেহেতু উভয় ক্ষেত্রে আসল (M), সময় (2 বছর) এবং সবৃদ্ধি মূলধন (N) একই,
সেহেতু যে পদ্ধতিতে বছরে বেশিবার মুনাফা গণনা করা হয় (Half-yearly), সেখানে কাঙ্ক্ষিত মুনাফা পেতে তুলনামূলক কম সুদের হার প্রয়োজন।
∴ R < S

Question: An agent sells goods worth Tk. 18,000. If his commission rate was 12.5​%, what was the amount of his commission?

Solution:

Commission = 12.5% of 18000
= (125/10) × (1/100) × 18000
= (125/1000) × 18000
= 2250

Question: If 5x + 7y = 41 and 7x + 5y = 43, what is the average of x and y?

Solution:
দেয়া আছে,
5x + 7y = 41 ...........(1)
7x + 5y = 43 ...........(2)

(1) ও (2) যোগ করে পাই, 
5x + 7y + 7x + 5y = 41 + 43
⇒ 12x + 12y = 84
⇒ 12(x + y) = 84
⇒ x + y = 84/12
∴ x + y = 7

∴ গড় = (x + y) /2
= 7/2
= 3.5

Question: A thief is noticed by a policeman from a distance of 200 m. The thief starts running and the policeman chases him. The thief and the policeman run at the rate of 10 km and 11 km per hour respectively. What is the distance between them after 6 minutes?

Solution:
Relative speed of the thief and policeman  =  (11 - 10) km/hr = 1 km/hr 
Distance covered in 6 minutes  = (1/60) × 6 km   = 1/10 km = 100 m
Therefore, Distance between the thief and policeman = (200 - 100) m = 100 m.

Question: A boat takes 12 hours to cover a certain distance upstream and 8 hours to cover the same distance downstream. What is the ratio between the speed of the boat and the speed of the water current, respectively?

Solution:
Let,
Upstream speed = x kmph
Downstream speed = y kmph

Since distance is same,
x × 12 = y × 8
⇒ 12x = 8y
⇒ y = 3x/2

Now,
Speed of boat = (y + x)/2
Speed of current = (y - x)/2

∴ Required ratio
= (y + x) : (y - x)
= [(3x/2) + x] : [(3x/2) - x]
= (5x/2) : (x/2)
= 5x : x
= 5 : 1

∴ The ratio of the speed of the boat to the speed of the water current is 5 : 1.

Question: A school has only four classes that contain 10, 20, 30 and 40 students respectively. The pass percentage of these classes are 20%, 30%, 60% and 100% respectively. Find the pass % of the entire school.

Solution:
Here,
10 of 20% = 2
20 of 30% = 6
30 of 60% = 18
40 of 100% = 40

The number of pass candidates are 2 + 6 + 18 + 40 = 66 out of total 100.

Hence, Pass percentage = 66%

Income from bank = 14% of Tk. 13200 = Tk. 1848
Number of shares purchased
= Tk. (13200/110)
= Tk. 120
Income from stock
= (15% of Tk. 100) × 120
= Tk. (15 × 20)
= Tk. 1800
∴ Loss = Tk. (1848 - 1800)
= Tk. 48