Bank Math

Question: At what angle the hands of a clock are inclined at 10 minutes past 4?

Solution:
Hours hand moves in 10 past 4 from 12 pm = (4 + 10/60) hour
= 4 + 1/6 
= 25/6 hour

We know,
Angle traced by hours hand in 12 hours =  360°

∴ Angle of hours hand in 25/6 hour,
= (360 × 25)/(12 ×6)
= 125°

Again,
Angle traced by Minutes hand in 60 min =  360°

∴ Angle of minutes hand in 10 min,
= (360 × 10)/60
= 60°

∴ Angle between hours and minutes hand  = (125° - 60°)
= 65°

Question: A boat travels 18 km downstream in 45 minutes. If the speed of the stream is 5 km/h, what is the speed of the boat in still water?

Solution:
স্রোতের অনুকূলে 45 মিনিটে যায় 18 কিমি
স্রোতের অনুকূলে 1 মিনিটে যায় 18/45 কিমি
স্রোতের অনুকূলে 1 ঘণ্টা বা 60 মিনিটে যায় (18 × 60)/45 কিমি
= 24 কিমি

∴ স্রোতের অনুকূলে বেগ = 24 কিমি/ঘণ্টা

দেওয়া আছে,
স্রোতের বেগ = 5 কিমি/ঘণ্টা।

∴ স্থির পানিতে নৌকার বেগ = স্রোতের অনুকূলে বেগ - স্রোতের বেগ
= 24 - 5 = 19 কিমি/ঘণ্টা।

Question: The marked price of a t-shirt was Tk. 800. A man bought the same for Tk. 420 after getting two successive discounts. the first being 25%. then the second discount rate is-

Solution:
Marked price = 800
Actual price = 420
First discount = 25%
Let the second discount be x%

Then, we can write

after 25% discount,
discounted amount = 800 × (25/100)
= 200

New price = 800 - 200
= 600 Tk

Again, second discount,
discounted amount = 600 × (x/100)
= 6x

New price = 600 - 6x

ATQ,
600 - 6x = 420
⇒ 600 - 420 = 6x
⇒ 6x = 180
⇒ x = 30

∴ Second discount = 30%

Quesiton: Find

Solution:

500 - 460 = 40, interest of two years
So, interest for three year is (40×3)/2 = 60 tk
∴ Initial investment = 460 - 60 = 400 tk

Question: Find the value of cos(2π/3).

Solution:
cos(2π/3)
= cos(π - π/3) 
= - cos(π/3)  ; [∵ (π - θ) দ্বিতীয় চতুর্ভাগে পড়ে এবং দ্বিতীয় চতুর্ভাগে cos ঋণাত্মক, তাই cos(π - θ) = - cos θ]
= - cos(60°)
= - 1/2

Let the two parts be x and (50 - x).
Then,
1/x + 1/(50 - x) = 1/12
(50 - x + x)/x(50 - x) = 1/12
x2 - 50x + 600 = 0
⇒ (x - 30)(x - 20) = 0
⇒ x = 30 or x = 20.
So, the parts are 30 and 20.

Question: The school has a total of 800 students. Yesterday, 12% of the boys and 18% of the girls were absent, and today 10% of the boys and 15% of the girls are absent. If today, 20 more students are present than yesterday, then find the total number of boys in the school.

Solution:
Let the Boys = B and the Girls = G
Then B + G = 800 ....... (1)

ATQ,
If absent 10% of boys and 15% of girls
So, present = 90% of boys and 85% of girls
⇒ 90% B + 85% G ........ (2)

If absent = 12% of boys and 18% of girls
So, present = 88% of boys and 82% of girls
⇒ 88% B + 82% G ......... (3)

Subtract equation (3) from (2)
90% B + 85% G - (88% B + 82% G) = 20
⇒ 2% B + 3% G = 20
⇒ 2B + 3G = 2000
⇒ 2B + 3(800 - B) = 2000 [From equation 1]
⇒ 2B - 3B = 2000 - 2400
∴ B = 400

Let the number be x
The problem could be 40 is subtracted from 60% of a number, the result is 50.
60% of x = 60/100×x= 60x/100
60x/100 - 40 = 50
(60x - 4000) /100 = 50
Cross multiplication
60x - 4000 = 50 ×100
60x -4000 = 5000
60x = 5000 + 4000
60x = 9000
x = 9000/60
x = 150
Hence, the number is 150

Work is done by 10 men in 1 day = 1/7
Work is done by 1 man in 1 day = (1/7)/10 = 1/70

Work is done by 10 women in 1 day = 1/14
Work is done by 1 woman in 1 day = 1/140

Work done by 5 men and 10 women in 1 day = 5 × (1/70) + 10 × (1/140)
= 5/70 + 10/140 = 1/7
∴ 5 men and 10 women can complete the work in 7 days.

Question: Two partners invest Tk. 1,20,000 and Tk. 80,000. After 6 months, they admit a new partner with Tk. 1,00,000. What is the ratio of their profits after one year?

Solution:
Profit sharing ratio depends on: Capital × Time.
Let the partners be A, B, and C.

A’s investment: Tk. 1,20,000 for 12 months
⇒ 1,20,000 × 12 = 14,40,000

B’s investment: Tk. 80,000 for 12 months
⇒ 80,000 × 12 = 9,60,000

C’s investment: Tk. 1,00,000 for 6 months
⇒ 1,00,000 × 6 = 6,00,000

Now, the ratio of profits:
14,40,000 : 9,60,000 : 6,00,000

Simplify = 12 : 8 : 5

∴ Ratio = 12 : 8 : 5

Relative speed = (60+40) km/hr
= 100×(5/18) m/sec
= 250/9 m/sec.
Distance covered in crossing each other
= (140+160)m= 300m
Required time
= 300×(9/250) sec
= 54/5 sec
= 10.8 sec

Question: Three times the middle of three consecutive even integers is 10 more than the smallest. What is the largest integer?

Solution: 
Let the three consecutive even integers are,
x – 2, x, x + 2. (where x = the middle integer)

According to the question,
Three times the middle is 10 more than the smallest.
∴ 3 × middle = smallest + 10
⇒ 3x = (x - 2) + 10
⇒ 3x = x - 2 + 10
⇒ 3x = x + 8
⇒ 3x - x = 8
⇒ 2x = 8
∴ x = 4

∴ The largest integer is x + 2 = 4 + 2 = 6.

Question: If SOS multiplied by TT equals TTTT, and if T = 9. Then what is the value of O?

Solution:
Given,
T = 9
∴ TT = 99
∴ TTTT = 9999

ATQ,
SOS × TT = TTTT
⇒ SOS × 99 = 9999
⇒ SOS = 9999/99
∴ SOS = 101

So, S = 1, O = 0, S = 1

We know the product of diagonals is 1/2×(product of diagonals)
Let one diagonal be d₁ and d₂
So as per question
1/2×d₁×d₂=150
1/2×10×d₂=150
d₂=150/5 = 30

Question: What is the slope of a line perpendicular to the line whose equation is 14x - 2y = 10?

Solution:
সরল রেখার সাধারণ সমীকরণ,
y = mx + c ......(1) (এখানে m = ঢাল)

যদি কোনো রেখার ঢাল হয় m, তবে তার লম্ব (perpendicular) রেখার ঢাল হবে,
m' = - (1/m)

এখন,
14x - 2y = 10
⇒ 2y = 14x - 10
⇒ y = (14/2)x - (10/2)
⇒ y = 7x - 5

(1) নং এর সাথে তুলনা করে পাই, m = 7

∴ লম্ব (perpendicular) রেখার ঢাল হবে, m' = - (1/7)

Question: Find the compound interest after 2 years on Tk. 14000 at the rate of interest 5% per annum.

Solution:
Let the sum P = 14000.
Rate of interest = 5%
Period = 2 years

Hence, compound interest = 14000(1 + 5/100)² - 14000
= 14000(1.05)² - 14000
= 14000 × 1.05 × 1.05 - 14000
= 15435 - 14000
= 1435

Let the distance be D km.
∴ Downstream Speed = D/4 km/hr
And Upstream Speed = D/5 km/hr
Given, Speed of current = 2 km/hr
Speed of the current = 1/2 ×(Downstream Speed - Upstream Speed)
2 = 1/2 ×(D/4 - D/5)
D = 80 km

Let the sum be Tk. x
Amount after 3 years on Tk. x at 20% per annum when interest is compounded annually
P(1 + R/100)^T
= x{1 + (20/100)}³
= x(120/100)³
= x(6/5)³
Compound interest = {x(6/5)³ - x}
= x{(6/5)³ - 1}
= x{(216/125) - 1}
= 91x/125
Simple interest PRT/100 = (x × 20 × 3)/100
= 3x/5
Given that difference between compound interest and simple interest is Tk. 48
(91x/125) - (3x/5) = 48
⇒ (91x - 75x)/125 = 48
⇒ 16x/125 = 48
⇒ x = (48 × 125)/16
= 3 × 125
= Tk. 375.

Cost of each share = (20 + 2.5% of 20) = tk.20.5
Therefore, number of shares = 8200/20.5 = 400

Question: The diagonals of a rhombus are 16 cm and 12 cm, in length. The side of the rhombus in length is:

Solution:
 
এখানে, রম্বসের কর্ণ দুটি পরস্পরকে সমকোণে সমদ্বিখণ্ডিত করে (90° কোণে)।
প্রথম কর্ণের অর্ধেক = 16/2 = 8 সে.মি.
এবং দ্বিতীয় কর্ণের অর্ধেক = 12/2 = 6 সে.মি.
এই অর্ধেক অংশ দুটি একটি সমকোণী ত্রিভুজের ভূমি ও লম্ব গঠন করে এবং রম্বসের বাহুটি হয় ত্রিভুজের অতিভুজ।

এখন, পিথাগোরাসের উপপাদ্য প্রয়োগ করে পাই, 
অতিভুজ² = ভূমি² + লম্ব²
অতিভুজ = √(ভূমি² + লম্ব²)
= √(8² + 6²)
= √(64 + 36)
= √(100)
∴ অতিভুজ = 10

সুতরাং, রম্বসের প্রতিটি বাহুর দৈর্ঘ্য 10 সে.মি.।

Product of two numbers = Product of their HCF and LCM.
Let one number = x
25 × x = 5 × 150
⇒ x = (5 × 150)/25
⇒ x = 30.

Question: Mr. Babul is aged three times than his son Mahi. After 6 years, he would be two and a half times of Mahi's age. After further 6 years, how many times would he be of Mahi's age?

Solution:
Let Mahi's present age be x years.
Then, Mr. Babul's present age = 3x years 

∴ After 6 years, Mahi's age will be (x + 6) years 
And, after 6 years, Mr. Babul's age will be (3x + 6) years

ATQ,
3x + 6 = (5/2) × (x + 6)
⇒ 3x + 6 = (5x + 30)/2  
⇒ 6x + 12 = 5x + 30
⇒ 6x - 5x = 30 - 12
⇒ x = 18

∴ After 6 years, Mahi's age will be (18 + 6) years
= 24 years
∴ After 6 years, Mr. Babul's age will be (3 × 18) + 6) years
= 54 + 6
= 60 years

∴ After further, 6 years, Mahi's age will be (24 + 6) years
= 30 years
∴ After 6 years, Mr. Babul's age will be (60 + 6) years
= 66 years

∴ Required Ratio = 66/30 
= 11/5 

∴ After further 6 years, Mr. Babul's age will be 11/5 times of Mahi's age.

A's share : B's share = Ratio of their 1 day's work
= 1/8 : 1/12
= 3 : 2.
∴ B's share = Tk. 200 × (2/5)
= Tk. 80.

Question: A sum of money doubles itself in 8 years at a certain rate of simple interest. In how many years will it become four times itself at the same rate of interest?

Solution: 
Given that, 
The sum doubles itself in 8 years.
Amount after 8 years = 2P
Simple Interest for 8 years = P

We know,
SI = (P × r × n)/100
⇒ P = (P × r × 8)/100
⇒ 8r = 100
⇒ r = 100/8
∴ r = 12.5% per year

Now, we want to find in how many years the sum becomes four times itself.
Amount = 4P
Interest needed = 4P - P = 3P

We know,
SI = (P × r × n)/100 
⇒ 3P = (P × 12.5 × n)/100.  ; [r = 12.5%]
⇒ 3 = (12.5 × n)/100
⇒ n = (3 × 100)/12.5
∴ n = 24 years

∴ The sum will become four times itself in 24 years.

Question: A cricketer scores 60 runs in the 10th innings and increases his average by 2 runs. What is his average after the 10th innings?

Solution:
ধরি, প্রথম 9 ইনিংসের গড় রান = x
∴ 9 ইনিংসের মোট রান = 9x

10 তম ইনিংসে 60 রান করার পর নতুন গড় = x + 2
সুতরাং, 10 ইনিংসের মোট রান = 10 × (x + 2)

এখানে, 9 তম ইনিংসের মোট রান এবং 10 তম ইনিংসের মোট রানের পার্থক্য = 60.
অর্থাৎ,
10(x + 2) - 9x = 60
⇒ 10x + 20 - 9x = 60
⇒ x + 20 = 60
∴ x = 40

অতএব, ১০ তম ইনিংসের পর গড় = 40 + 2 = 42

Let total quantity of original milk = 1000 gm
Milk after first operation = 80% of 1000 = 800 gm
Milk after second operation = 80% of 800 = 640 gm
Milk after third operation = 80% of 640 = 512 gm
∴ Strength of final mixture = 51.2%

Time taken by one tap to fill half of the tank = 3 hrs.
Part filled by the four taps in 1 hour = 4 × (1/6) = 2/3
Remaining part = (1 - 1/2) = 1/2
2/3 : 1/2 :: 1 : x
=> x = (1/2 × 1 × 3/2) = 3/4
So, total time taken = 3 hrs. 45 mins.

Question: Five times a whole number is equal to three less than twice the square of the number. Find the number?

Solution: Let the required whole number be x.

According to the question,
5x = 2x² – 3
⇒ 2x² – 5x – 3 = 0
⇒ 2x² - 6x + x - 3 = 0
⇒ 2x(x - 3) + 1(x - 3)
⇒ (2x + 1)(x – 3) = 0

So,
2x + 1 = 0 or x – 3 = 0
⇒ x = –1/2 or x = 3

Since x must be a whole number,
∴ the required number is 3.

Let A, B, C represent their respective weights. Then, we have:
A + B + C = (45 x 3) = 135 .... (i)
A + B = (40 x 2) = 80 .... (ii)
B + C = (43 x 2) = 86 ....(iii)
Adding (ii) and (iii), we get: A + 2B + C = 166 .... (iv)
Subtracting (i) from (iv), we get : B = 31.
B's weight = 31 kg.

Let the numbers be x and y
According to question,(x>y)
x²−y²=45
(x+y)(x−y)=45
Make factor of 45
15×3
9×5
45×1
Total 3pairs
These pairs gives the value of x and y which satisfy the given condition.

Question: A man takes 6 hours to walk to a place and ride back to the starting point. If he were to walk both ways, it would take him 9 hours. How much time would he take to ride both ways?

Solution:
Time taken in walking both ways = 9 hours ................(i)
Time taken in walking one way and riding back = 6 hours ........(ii)

By the equation (ii) × 2 - (i), we have:
Time taken by the man in riding both ways
= (6 × 2) - 9
= 12 - 9
= 3 hours

∴ The time taken by him to ride both ways is 3 hours.

Question: Fedora Technologies acquired 60 hard-drives and tried selling them at BDT 2400 apiece at 20 percent markup on cost price. After 3 months, 6 remained unsold and were returned to manufacturer at 50 percent refund of cost. Find out Fedora's approximate profit margin as a percentage of initial acquisition cost of the 60 hard-drives.

Solution:
ধরি,
প্রতি ইউনিটের ক্রয়মূল্য = x

Markup ২০%, অর্থাৎ,
বিক্রয়মূল্য = 1.2x = 2400
⇒ x = 2400/1.2​
= (2400 × 10)/12
= 2000 টাকা

অতএব,
প্রতি হার্ডড্রাইভের ক্রয়মূল্য = ২০০০ টাকা
∴ মোট ক্রয়মূল্য = 60 × 2000 = 120000 টাকা

আবার,
বিক্রিত হার্ডড্রাইভ = 60 - 6 = 54
প্রতি হার্ডড্রাইভ বিক্রয়মূল্য = 2400 টাকা
∴ মোট বিক্রয়মূল্য =54 × 2400 = 129600

অবিক্রিত 6টি ফেরত দিয়ে,
রিফান্ড = 6 × (50% × 2000)
= 6 × 1000
= 6000 টাকা

∴ মোট আয় = (বিক্রয় + রিফান্ড)
= 129600 + 6000
= 135600 টাকা

∴  লাভ = 135600 - 120000 = 15600 টাকা

∴  লাভের হার = (15600/120000) × 100 = 13%