Bank Math

Question: If m and n are even numbers, which of the following is always even?

Solution:
Let m = 2 and n = 4 (both are even numbers)
a) m + n + 1 = 2 + 4 + 1 = 7 ......... Odd
b) mn + 3 = (2 × 4) + 3 = 8 + 3 = 11 ......... Odd
c) 2m + n = (2 × 2) + 4 = 4 + 4 = 8 ......... Even
d) m² + n + 1 = (2)² + 4 + 1 = 4 + 4 + 1 = 9 ......... Odd

Answer: c) 2m + n is always even

Let the sum be Tk.P.then
P(1+R/100)³=6690…(i) and
P(1+R/100)⁶=10035…(ii)
On dividing,we get (1+R/100)³=10025/6690=3/2.
Substituting this value in (i),we get:
P×(3/2)=6690 or
P=(6690× 2/3)=4460
Hence,the sum is Tk.4460.

Question: What will come in the place of the question mark in the following question- 40% of 50% of 3/4 of 1800 = ?

Solution:
40% of 50% of 3/4 of 1800 = ?
⇒ 40% × 50% × (3/4) × 1800 = ?
⇒ (40/100) × (50/100) × 3/4 × 1800 = ?
⇒ (2/5) × (1/2) × (3/4) × 1800 = ?
⇒ (6/40) × 1800 = ?
⇒ ? = 270

∴ The question mark is replaced by 270

1 থেকে 70 পর্যন্ত মোট সংখ্যা 70 টি।
এর মধ্যে এককের স্থানে 1 অথবা 9 আছে এমন সংখ্যা: 1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69 মোট 14 টি
∴ সংখ্যা হার = (14/70 × 100) % = 20 %

Question: If 2^{x + 1} = 16, then x = ?

Solution: 
Given that, 
2^{x + 1} = 16
⇒ 2^{x + 1} = 2⁴
⇒ x + 1 = 4
⇒ x = 4 - 1
∴ x = 3

So the value of x is 3

Question: Some principal becomes Tk. 1,560 as profit-principal in 3 years and Tk. 1,920 as profit-principal in 6 years. Find the rate of profit.

Solution: 
Given,
Principal + Profit for 3 years = 1560
Principal + Profit for 6 years = 1920

So the profit of 3 years I = Tk. (1920 - 1560)
= Tk. 360

So the principal P = Tk. (1560 - 360)
= Tk. 1200

∴ The rate of profit r = (I × 100)/(P × n)
= (360 × 100)/(1200 × 3)
= 10

Question: If , what is the value of A?

Solution:

Question: Gold is 19 times as heavy as water and copper is 9 times as heavy as water. In what ratio should these be mixed to get an alloy 10 times as heavy as water? 

Solution: 
let, gold is 19x tme heavy and copper 9y times heavy as water 

19x + 9y = 10 (x + y)
⇒ 19x + 9y = 10x + 10y 
⇒ 19x - 10x = 10y - 9y
 ⇒ 9x = y

∴ x/y = 1/9
= 1 : 9 

Question: If a 20-meter tall pole creates a shadow of length 20√3 meters, what is the angle of elevation of the sun?

Solution:
 
খুঁটির উচ্চতা AB = 20 m
খুঁটির ছায়ার দৈর্ঘ্য BC =20√3 m

ΔABC হতে পাই,
tanθ = লম্ব/ভূমি
বা, tanθ = AB/BC
বা, tanθ = 20/(20√3)
বা, tanθ = 1/√3
বা, tanθ = tan30°
∴ θ = 30°

Radius of incircle = a/2√3.
Area of incircle = (π × a²)/12 cm²
∴ πa²/12 = 154
⇒ a² = (154 × 12 × 7)/22
⇒ a = 14√3
∴ Perimeter of the triangle = (3 × 14√3) cm
= (14 × 1.732) cm
= 72.7 cm (approx.)

Question: Mr. Rahim deposited a certain amount of money in a bank. Upon maturity, he received a total of Tk. 72,000, where the ratio of interest to principal was 1 : 5. If the simple interest rate was 4%, for how many years was the money invested?

Solution:
প্রদত্ত তথ্য অনুযায়ী, আসল (P) এবং সুদের (I) অনুপাত = 5 : 1
মোট প্রাপ্ত টাকা (সুদ-আসল) = 72,000 টাকা
অনুপাতের যোগফল = 5 + 1 = 6

আসল (P) = 72,000 এর (5/6) অংশ = 60,000 টাকা
সুদ (I) = 72,000 এর (1/6) অংশ = 12,000 টাকা

এখানে,
I = 12,000 টাকা
P = 60,000 টাকা
R = 4%
T = ?

আমরা জানি, SI = (P × R × T)/100
⇒ 12,000 = (60,000 × 4 × T)/100
⇒ 12,000 = 600 × 4 × T
⇒ 12,000 = 2,400 × T
⇒ T = 12,000/2,400
⇒ T = 5

∴ টাকাটি 5 বছরের জন্য বিনিয়োগ করা হয়েছিল।

Question: A particle moves such that its displacement is described by S = 10t - t², what is its displacement after 2 seconds?

Solution:
Given that,
S(t) = 10t - t²

We want S at t = 2 seconds.
Now, 
S(2) = 10 × 2 - 22 = 20 - 4
∴ S(2) = 16

So the displacement after 2 seconds is 16.

Question: In January, the value of a stock increased by 50%; and in February, it decreased by 20%. In March, it increased by 25%; and in April, it decreased by 10%. If a person invested Tk. 200 in the stock on January 1 and sold it at the end of April, what was the percentage change in the price of the stock?

Solution:
At the end of January,
The value of the stock is = Tk. 200 + 50% of (Tk. 200)
= Tk. 200 + Tk. 100 = Tk. 300.

At the end of February,
The value of the stock is = Tk. 300 - 20% of (Tk. 300)
= Tk. 300 - Tk. 60 = Tk. 240.

At the end of March,
The value of the stock is = Tk. 240 + 25% of (Tk. 240)
= Tk. 240 + Tk. 60 = Tk. 300.

At the end of April,
The value of the stock is = Tk. 300 - 10% of (Tk. 300)
= Tk. 300 - Tk. 30 = Tk. 270.

Now, the percentage change in price is,
= (Change in price/Original price) × 100%
= (270 - 200)/200 × 100%
= (70/200) × 100%
= 35%

Question: If the sum of the 5th term and the 15th term of an arithmetic progression is 20. What is the sum of the first 19 terms of the progression?

Solution:
ধরি, সমান্তর ধারার প্রথম পদ a এবং সাধারণ অন্তর d
আমরা জানি,
সমান্তর ধারার, n তম পদ = a + (n - 1)d
এবং n তম পদের যোগফল: S_n = n/2[2a + (n - 1)d]

∴ 5 তম পদ = a + 4d 
15 তম পদ = a + 14d

প্রশ্নমতে,
(a + 4d) + (a + 14d) = 20
⇒ 2a + 18d = 20

∴ প্রথম 19 পদের এর যোগফল:
S₁₉ = 19/2 [2a + 18d]
= 19/2 × 20
= 19 × 10
 = 190

Question: How many bricks are required to build a wall that is 8 m long, 6 m high, and 11 cm thick, if each brick measures 25 cm × 60 cm × 11 cm?

Solution:
Given,
Wall long = 8 m = 800 cm
Wall thick = 11 cm
Wall high = 6 m = 600 cm

∴ Volume of the wall = (800 × 600 × 11) cm³

∴ Volume of the brick = 25 cm × 11 cm × 60 cm

∴ bricks need to build the wall = Volume of the wall ÷ Volume of the brick
= (800 × 600 × 11) ÷ (25 × 11 × 60)
= 320

Question: A bag contains 6 blue marbles, 4 green marbles, and 5 black marbles. If one marble is drawn at random from the bag, what is the probability that the marble is green?

Solution:
Total number of marbles = 6 + 4 + 5 = 15
Number of green marbles = 4

Therefore, P(green) = Number of green marbles/Total marbles
= 4/15

So the probability is 4/15.

Question: Two pipes can fill a tank in 8 and 12 minutes respectively and a waste pipe can empty 4 gallons per minute. All the three pipes working together can fill the tank in 24 minutes. The capacity of the tank is:

Solution:
Work done by the waste pipe in 1 minute
= (1/24) - [(1/8) + (1/12)]
= (1 - 3 - 2)/24
= - 1/6 [ - ve sign means emptying ]

∴ Volume of (1/6) part = 4 gallons
Volume of whole
= (6 × 4) gallons
= 24 gallons

Question: If
​
​Solution:
​

Question: If two coins are tossed, what is the probability of getting at most one head?

Solution:
দুটি মুদ্রা নিক্ষেপ করলে সম্ভাব্য ফলাফলগুলো হলো:
S = {HH, HT, TH, TT}
এখানে, মোট ফলাফল সংখ্যা = 4

"at most one head" মানে সর্বোচ্চ একটি head অর্থাৎ 0টি head অথবা 1টি head.
0টি head এর ক্ষেত্রে: TT (1টি)
1টি head এর ক্ষেত্রে: HT, TH (2টি)
∴ অনুকূল ফলাফল n(E) = 1 + 2 = 3টি

সুতরাং, সর্বোচ্চ একটি head পাওয়ার সম্ভাবনা = n(E)/n(S)
= 3/4

A : B = 1000 : (1000 - 100)
= 1000 : 900
= 10 : 9
B : C = 1000 : (1000 - 60)
= 1000 : 940
100 : 94
= 9 : 846/100
Therefor, A : C = 10 : 846/100
= 10000 : 846
hence A can beat C by = (10000 - 846)
= 154 metres.

Question: What is the value of cos 120°?

Solution: 
cos 120°
= cos(90° + 30°)
= - sin 30°
= - 1/2

Question: A 314 cm long copper strip is bent into a round wheel. What is the wheel’s diameter?

Solution:
Length of the strip = Circumference of the wheel = 314 cm

We know,
Circumference C = π × d

wheel’s diameter, d = C ÷ π = 314 ÷ 3.14 = 100 cm

Question: What will be the number in the question mark?
3, 7, 15, 27, 43, ?

Solution:
প্রদত্ত ধারাটি হলো: 3, 7, 15, 27, 43, ?
ধারার সংখ্যাগুলোর মধ্যে পার্থক্য নির্ণয় করি:
7 - 3 = 4
15 - 7 = 8
27 - 15 = 12
43 - 27 = 16

এখানে, প্রতিবার পার্থক্য 4 করে বৃদ্ধি পাচ্ছে।
∴ পরবর্তী পার্থক্য হবে = 16 + 4 = 20
∴ পরবর্তী সংখ্যাটি হবে = 43 + 20 = 63

অতএব, প্রশ্নবোধক স্থানে 63 বসবে।

Shortcut: 3 (+4)→ 7 (+8)→ 15 (+12)→ 27 (+16)→ 43 (+20) → 63.

Question: A cricket team consists of 11 players. The captain is 24 years old, and the wicketkeeper is 1 years older than the captain. When these two players are excluded, the average age of the remaining players becomes one year less than the average age of the whole team. What is the average age of the team?

Solution:
Let,
the average age of the whole team is x years.
⇒ 11x - (24 + 25) = 9(x - 1)
⇒ 11x - 49 = 9x - 9 
⇒ 11x - 9x = 40
⇒ 2x = 40
⇒ x = 20.

So, the average age of the team is 20 years.

Question: If a 40√3 meter ladder is placed against a 60 meter wall such that it just reaches the top of the wall, the angle of elevation of the wall is -

Solution:

এখানে,
দেয়ালের উচ্চতা (লম্ব) = 60 মিটার
মইয়ের দৈর্ঘ্য (অতিভুজ) = 40√3 মিটার
উন্নতি কোণ (angle of elevation) = θ

আমরা জানি,
sinθ = লম্ব/অতিভুজ
⇒ sinθ = 60/40√3
⇒ sinθ = 3/2√3
⇒ sinθ = √3/2
⇒ sin θ = sin60°
∴ θ = 60°

সুতরাং, দেয়ালের উন্নতি কোণ ((angle of elevation) = 60°

Question: Speed ​​of the boat and current are 12 and 4 km/h respectively. How much time will it take for the boat to travel 64 km downstream and then return the same distance upstream?

Solution:
We know,
Effective speed with the current = Actual speed + speed of stream
= (12 + 4)  km/h
= 16 km/h

∴ Time taken to cover 64 km = (64 ÷ 16) hours 
= 4 hours

Effective speed against the current = Actual speed - speed of stream
= (12 - 4) km/h
= 8 km/h

∴  Time taken to return 64 km = (64 ÷ 8) hours
= 8 hours

∴ Total time taken = (4 + 8) hours 
= 12 hours

CP× (76/100) = 912
CP = 12 × 100
=> CP = 1200

Question: The perimeter of a rectangle is 30cm. If the breadth of the rectangle is 6cm, then the ratio of the length and breadth will be-

Solution: 
Let the length be x cm.

Now
2(6 + x) = 30 
⇒ 6 + x = 15
⇒ x = 15 - 6
⇒ x = 9
∴ Length = 9 cm

∴ Required ratio = Length : Breadth
= 9 : 6
= 3 : 2

Question: In your wallet, there are Tk 1000, Tk 500, and Tk 100 notes in the ratio 3 : 5 : 2, amounting to Tk 22,800. Find the number of each note respectively.

Solution:
Let,
The number of Tk 1000 notes is 3x
The number of Tk 500 notes is 5x
The number of Tk 100 notes is 2x

ATQ,
1000 × 3x + 500 × 5x + 100 × 2x = 22800
⇒ 3000x + 2500x + 200x = 22800
⇒ 5700x = 22800
⇒ x = 22800 / 5700
⇒ x = 4

Number of Tk 1000 note = 3x = 3 × 4 = 12
Number of Tk 500 note = 5x = 5 × 4 = 20
Number of Tk 100 note = 2x = 2 × 4 = 8

Therefore, the number of Tk 1000, Tk 500, and Tk 100 notes are respectively 12, 20, and 8.

Question: A mechanic charges a fixed service fee of Tk. 80, along with Tk. 40 for each hour of labor. If a customer’s total bill must not exceed Tk. 320, determine the maximum number of full hours the mechanic can work.

Solution:
Given that,
Fixed service fee = Tk. 80
Charge per hour = Tk. 40
Maximum total bill = Tk. 320

Let h = number of full hours the mechanic works.

Now, total cost equation,
80 + 40h ≤ 320
⇒ 40h ≤ 320 - 80
⇒ 40h ≤ 240
⇒ h ≤ 240/40
⇒ h ≤ 6

So, the mechanic can work a maximum of 6 full hours.

ATQ,
Cost price = 100/150 × 6000 = 4000
Selling price = 4000+ 500 = 4500
So, Percentage needed to be above the cost price by = 500/4000 × 100 = 12.5%

Previous speed = x
A/Q,
32/x - 32/(x + 4) = 4
Or, 32x + 128 - 32x/x(x + 4) = 4
Or, (x + 4)x = 32
Or, x² + 4x - 32 = 0
Or, x² + 8x - 4x - 32 = 0
Or, (x + 8) (x - 4) = 0
So, the previous speed was 4 kmph.
And, Present speed will be 4 + 4 = 8 kmph.

According to question,        
XY/100 + YX/100
= 2XY/100
= 2% o fXY

Let the length of the tunnel = x meter
Then, distance = (800 + x) meter.
Time = 1 minute = 60 seconds.
Speed = 78 km/hr
= 78 × (5/18)
= (65/3) m/s
According to the question,
800 + x = 60 × (65/3)
⇒ 800 + x = 1300
⇒ x = 500 meter.

Question: A man was asked to state his age in years. His reply was, "Take my age 4 years hence, multiply it by 4 and then subtract 4 times my age 4 years ago, and you will know how old I am". What is the age of the man?

Solution:
Let the age of the man be = p years
Then,
⇒ 4(p + 4) - 4(p - 4) = p
⇒ 4p + 16 - 4p + 16 = p
⇒ p = 32
∴ The age of the man is 32 years.

Given that,
P = Rs 50,000
R = 3.5%
T = 3 years

S.I.
= (P × R × T)/100
= (50,000 × 3.5 × 3)/100
= Tk. 5250.