Bank Math

Question: In a factory, 10 machines can produce 10 toys in 10 minutes. How many minutes will it take for one machine to produce one toy?

Solution:
10টি মেশিন 10টি খেলনা তৈরি করে 10 মিনিটে।
1টি মেশিন 10টি খেলনা তৈরি করে = 10 × 10 মিনিটে 
1টি মেশিন 1টি খেলনা তৈরি করে = (10 × 10)/10 মিনিটে 
= 10 মিনিটে।

Age of Raju when he got the job = 17 + 3 + 2 + 1 = 23 yr

Age of Raju at the time of marriage = 23 + 3 = 26 yr

Question: In a certain code, WATER = 41253 and CROWD = 67840, how is TOWER coded in the language?

Solution:
Given,
W  A  T  E  R
↓    ↓  ↓  ↓  ↓
4   1  2  5  3

and
C  R  O  W  D
↓  ↓   ↓   ↓   ↓
6  7  8   4   0

So
T  O  W  E  R
↓  ↓   ↓   ↓   ↓
2  8  4   5   3

Question: If 2^{x - 6} = 1/32, the value of x/2 is:


Solution:

Question: A container has 64 liters of milk. From this container, 16 liters of milk was taken out and replaced with water. This process was repeated twice. What is the final ratio of milk and water in the container? 

Solution:
Initial quantity of milk = 64 liters
Removed quantity = 16 liters
Total quantity = 64 liters
Number of times the process is repeated = 2

Remaining milk = Initial quantity × {1 - (Removed Quantity/Total Quantity)}ⁿ
= 64 × {1 - (16/64)}²
= 64 × (3/4)²
= 36 liters

So, Final quantity of milk = 36 liters
Final quantity of water = (64 - 36) = 28 liters

Therefore, Ratio of milk to water = 36/28 = 9 : 7

The average of 11 results = 60
∴ The total of 11 results = 60 × 11 = 660
Average of first six results = 58
∴ Total of first six results = 58 × 6 = 348
Average of last six results = 63
∴ Total of last six results = 63 × 6 = 378
∴ sixth results = total of first and last six results - total of 11 results 
                     = (348 + 378) - 660
                     = 726 - 660
                     = 66
Answer: 66

Question: 


Solution:

Question: If the simple interest on Tk. N at N% per annum for 5 years amounts to Tk. 2N, what is the value of N?

Solution:
Given,
Principal, P = N
Rate, r = N% = N/100
Time, n = 5 years
Simple Interest, I = 2N

We know,
I = (P × r × n)/100
⇒ 2N = N × (N/100) × 5
⇒ 2N = 5N²/100
⇒ 2N × 100 = 5N²
⇒ 200N = 5N²
⇒ N² = 40N
⇒ N = 40

∴ The value of N = 40

Question: If xsinA = 1, xcosA = √3, then the value of (√3tanA + 2) = ?

Solution:
Given that
xsinA =1, xcosA = √3

xsinA/xcosA = 1/√3
⇒ tanA = 1/√3
⇒ √3tanA = 1
⇒ √3tanA + 2 = 1 + 2
∴ √3tanA + 2 = 3

Question: The population of a town increases every year by 5%. If its present population is 60,000, then after 2 years, what will be the population?

Solution: 
We know, Population after n years = P × [1 + (r/100)]ⁿ

∴ Population after 2 years = 60000 × [1 + (5/100)]²  
= 60000 × (1 + 0.05)² 
= 60000 × 1.1025
= 66150

1 hectare = 10,000 m²
So, Area = (1.5 x 10000) m² = 15000 m².
Depth =5/100 m=1/20m.
Volume = (Area x Depth) =15000 x1/20m³= 750 m^3.

ধরি, price x টাকা
TV এর price বাড়ার পর = x + x এর 20%
= 1.2x
Computer এর price কমার পর = x - x এর 10%
= 0.9x
পূর্বে 4 টি TV এবং 4 টি computer এর price = 8x
এখন 4 টি TV এবং 4 টি computer এর price = 4.8x + 3.6x
= 8.4x
দাম বেশি হয় = (8.4x - 8x) = 0.4x টাকা

সুতরাং, খরচ বাড়বে = (0.4/8) × 100
= 5%

Question: A rectangular sheet of paper, 10cm long and 8cm wide has squares of side 2cm cut from each of its corner. The sheet is then folded to form a tray of depth 2cm. What is the volume of this tray?

Solution: 
Length of tray = 10 - (2 × 2) = 10 - 4 = 6 cm.
Breadth of tray = 8 - (2 × 2) = 4 cm.
Depth of tray = 2 cm.
∴ Volume of tray = 6 × 4 × 2 = 48 cm³

Question: Solve the inequality, (x/4) < (4x - 1)/15

Solution: 
Given that, 
x/4 < (4x - 1)/15
⇒ 15x < 16x - 4
⇒ 15x - 16x < - 4
⇒ - x < - 4
⇒ x > 4 

Question: If one-fourth of one-sixth of a number is 5, then what is 3/4 of the number?

Solution:
Let the number be x.
According to the question,
(1/4) of (1/6) of x = 5
⇒ (1/4) × (1/6) × x = 5
⇒ (1/24)x = 5
⇒ x = 5 × 24
∴ x = 120

Now,
(3/4) of x = (3/4) × 120
= 360/4
= 90

Question: The radius of circle A is r, and the radius of circle B is 2r/3. What is the ratio of the area of circle A to the area of circle B?

Solution:
The radius of circle A = r
The area of circle A = πr²

The radius of circle B = 2r/3
The area of circle B = π(2r/3)² = 4πr²/9

∴ The ratio of the area of circle A to the area of circle B = πr² : 4πr²/9
= 1 : 4/9
= 9 : 4

Question: A and B together can complete a work in 10 days. A alone can complete it in 30 days. If B works only for half a day daily, then in how many days will A and B together complete the work?

Solution:
দেওয়া আছে,
A ও B একসাথে কাজটি সম্পন্ন করে = 10 দিনে
সুতরাং, তাদের এক দিনের কাজ = 1/10 অংশ

A একা কাজটি সম্পন্ন করে = 30 দিনে
সুতরাং, A এর এক দিনের কাজ = 1/30 অংশ

অতএব, B এর এক দিনের কাজ = (A ও B এর একসাথে কাজ) - (A এর একা কাজ)
= 1/10 - 1/30 অংশ
= (3 - 1)/30 অংশ
= 2/30 অংশ
= 1/15 অংশ

B প্রতিদিন অর্ধেক দিন কাজ করলে, তার প্রতিদিনের কাজ হবে,
= (1/15)/2 অংশ
= 1/30 অংশ

এখন, A (পুরো দিন) এবং B (অর্ধেক দিন) একসাথে কাজ করলে তাদের প্রতিদিনের মোট কাজ হবে:
= (A এর এক দিনের কাজ) + (B এর প্রতিদিনের কাজ)
= 1/30 + 1/30 অংশ
= 2/30 = 1/15 অংশ

যেহেতু তারা প্রতিদিন কাজের 1/15 অংশ সম্পন্ন করে, তাই সম্পূর্ণ কাজটি সম্পন্ন করতে তাদের সময় লাগবে 15 দিন।

সুতরাং, তারা একসাথে 15 দিনে কাজটি সম্পন্ন করবে।

Question: A man sells a watch at a 5% loss. If he had sold it for Tk. 56.25 more, he would have made a 10% profit. What was the cost price of the watch?

Solution:
Let,
the price of watch x taka

According to the question,
x(100% - 5%) + 56.25 = x(100% + 10 %)
→ 95%x + 56.25 = 110% x
→ 15%x = 56.25
→ x = (5625/15)
∴ x = 375 taka

∴ the cost of the watch is 375 taka.

Question: In a trapezoid, the lengths of the two parallel bases are 8 cm and 16 cm. If the height of the trapezoid is 6 cm, find the area of the trapezoid.

Solution:
Given that,
Trapezoid with bases a = 8 cm and b = 16 cm
Height, h = 6 cm

We know,
Area of trapezoid = (1/2) × (sum of bases) × height
= (1/2) × (a + b) × h
= (1/2) × (8 + 16) × 6
= (1/2) × 24 × 6
= 12 × 6
= 72

∴ The area of the trapezoid is 72 sq. cm

Question: The supplement of an angle exceeds twice the angle by 30°. Then the angle is equal to-

Solution:
Let the angle be x
Then, its supplement = 180 - x

According to the question,
180 - x = 2x + 30
⇒ 180 - 30 = 3x
⇒ 150 = 3x
⇒ x = 50°

Question: In a 90-liter mixture of milk and water, the ratio of milk to water is 2 : 1. How many liters of water must be added to make the ratio become 1 : 2?

Solution:
Total mixture = 90 litres
Given ratio (milk : water) = 2 : 1

Milk = 90 × (2/3) = 60 litres
Water = 90 − 60 = 30 litres

To make the ratio 1 : 2, x liters of water need to be added.
milk : water = 60 : (30 + x)

So,
60 / (30 + x) = 1 / 2

Cross-multiplying,
2 × 60 = 30 + x
120 = 30 + x
x = 120 − 30
∴ x = 90

Quantity of water to be added = 90 litres.

Question: In how many ways can the letters of the word "EQUATION" be arranged such that the consonants occupy only the even positions?

Solution:
এখানে "EQUATION" শব্দটিতে মোট বর্ণ আছে 8টি।
ব্যঞ্জনবর্ণ (Consonant) আছে 3টি: Q, T, N
স্বরবর্ণ (Vowel) আছে 5টি: E, U, A, I, O
8টি বর্ণের মধ্যে জোড় স্থান (Even positions) আছে 4টি (2nd, 4th, 6th এবং 8th)।

4টি জোড় স্থানের মধ্যে 3টি ব্যঞ্জনবর্ণ সাজানোর উপায় = ⁴P₃ = 24
বাকি (8 - 3) = 5টি স্থানে 5টি স্বরবর্ণ সাজানোর উপায় = 5! = 120

∴ ব্যঞ্জনবর্ণগুলোকে কেবল জোড় স্থানে রেখে মোট বিন্যাস সংখ্যা = 24 × 120
= 2,880

অতএব, EQUATION শব্দটির ব্যঞ্জনবর্ণগুলোকে জোড় স্থানে রেখে মোট 2,880 উপায়ে সাজানো যাবে।

15min = 1/4hrs
1 hr → 4 kms
1/4hr → 4/4 kms
So, length of the bridge= 1 km = 1000 metres

Question: Find the equation of the line with x-intercept = 4 and y-intercept = 3.

Solution:
Given, x-intercept = 4,
So, the line passes through (4, 0).
y-intercept = 3,
So, the line passes through (0, 3).

We know,
The intercept form of a line is:
(x/a) + (y/b) = 1, where a = x-intercept, b = y-intercept.
⇒ (x/4) + (y/3) = 1
⇒ (3x + 4y)/12 = 1
⇒ 3x + 4y = 12
⇒ 3x + 4y - 12 = 0

∴ The equation of the line is 3x + 4y - 12 = 0

tanθ = লম্ব/ভূমি
[এখানে, লম্ব = খুঁটির দৈর্ঘ্য এবং ভূমি = ছায়ার দৈর্ঘ্য] 
⇒ tanθ = 6/2√3
⇒ tanθ = tan 60°
∴ θ = 60°

Question: If log_X(81/16) = -4, what is the value of X?

Solution:
By definition of logarithm:
log_X(y) = n
⇒ Xⁿ = y

Here,
log_X(81/16) = - 4
 ⇒ X^-4 = 81/16
⇒ X⁴ = 16/81
⇒ X = ⁴√(16/81)
⇒ X = 2/3

We know that
The ratio of Investment x Time = Ratio of Profit
∴ (A's investment × Time) : (B's investment × Time) = Profit of A : Profit of B

∴ Tk. 8000 × 12 months : Tk. 16000 × 9 months : Tk. 40000 × ? months = 6 : 9 : 5
∴ 96000 : 144000 : 40000 × ? = 6 : 9 : 5
By direct observation we can say, if common factor is K, then
6K = 96000;
∴ K = 16000;
and 5K = 40000 × ?
? = 5k/40000
= (5 × 16000)/40000
= 2 months.

Here,
Total results S = {1, 2, 3, 4, .....19, 20}.
Let E = event of getting a multiple of 3 or 5 = {3, 6, 9, 12, 15, 18, 5, 10, 20}
∴ P(E) = n(E)/n(S)
= 9/20
= 4.5/10
= 0.45
Answer: 0.45

In the first minute the monkey climbs 1 meter.
In the second minute it slips 1/2 meter.
For every two minute it climbs 1/2 meter.

So Average speed = 1 meter/4 minutes
For 11 meters, time taken = 44 minutes.

For the last 1 meter jump add 1 minute.
So time taken = 45 minutes.

Question: A tank was 20% full of oil. The oil was poured into an empty 100-liter bucket, filling it halfway. What is half of the tank’s total capacity (in liters)?

Solution:
Let the total capacity of the tank = T liters.
The tank was 20% full 
now, oil volume = 20% of T = 0.2 × T

The oil fills 50% of the bucket
50% of 100 liters = 50 liters

So, 0.2 × T = 50 → T = 50 / 0.2 = 250 liters

Half of the tank = 250 ÷ 2 = 125 liters.

Question: Two trains are moving in opposite directions at speeds of 60 km/h and 90 km/h. Their lengths are 800 m and 700 m respectively. Find the time taken to cross each other.

Solution:
Relative speed = (60 + 90) km/h
= 150 × (5/18) m/sec
= 125/3 m/sec

Distance covered = (800 + 700) m
= 1500 m

Required time
= 1500 ÷ (125/3) sec
= (1500 × 3)/125 sec
= 36 sec

∴ The required time is 36 seconds.

প্রশ্ন: A train is going at 1/3 of its usual speed and it takes an extra 30 minutes to reach its destination. Find its usual time to cover the same distance.

সমাধান: 
Let,
The usual speed is x 
The distance is d 
The usual time is t min 

So, 
In usual speed, d = xt 

In 1/3 of usual speed, d = (x/3) × (t + 30) 

Now, We can say that,
xt = x(t + 30)/3 
⇒ t = (t + 30)/3
⇒ 3t = t + 30 
⇒ 2t = 30
∴ t = 15 

∴ The usual time to cover the same distance is 15 min.

Remaining apples (100 - 40) % = 60%
60% = 420 
∴ 100% = 420 × 100 / 60 = 700 

Let the number be 10x + 3
ATQ,
7(x + 3) = 10x + 3
⇒ 7x + 21 = 10x + 3
⇒ 21 - 3 = 10x – 7x
⇒ 3x = 18
⇒ x = 6
∴ The number is, 10×6 + 3 = 63

Take work done = 1
Let number of sailors who could not board = S.
So sailors who boarded = 2500 - S
∴ 2500 sailors x 40 days x 1 = (2500 - S) sailors x 50 days x 1
∴ S = 500 = Sailors who could not board the ship.

[Men = M; Days = D; Time/Hours = T; Work = W
M₁D₁T₁W₂ = M₂D₂T₂W₁
Note that - W₂ is on left side and W₁ is on right side]

প্রশ্ন: Haris and Sunny share some sweets in the ratio of 7 : 5. Haris has 12 more sweets than Sunny. How many sweets were there altogether?

সমাধান: 
Let,
Haris has 7x sweets 
Sunny has 5x sweets 
∴ Total sweets 7x + 5x = 12x

ATQ,
7x - 5x = 12
⇒ 2x = 12
∴ x = 6

∴ There were 12 × 6 = 72 sweets altogether.