Bank Math

Question: If 5 identical machines, operating at a constant speed, can manufacture 200 pencils in one minute, how many pencils will 12 such machines produce in 10 minutes at the same rate?

Solution:
Given,
In 1 minute, 5 machines can produce 200 pencils
In 1 minute, 1 machine can produce 200/5 = 40 pencils

So, in 10 minutes, 12 machines can produce = (40 × 12 × 10) pencils
= 4800 pencils

Question: The average of 5 numbers is 24. If 3 more numbers, with an average of 18 are added to these numbers, what will be the average of the combined 8 numbers?

Solution:
The average of 5 numbers is 24
∴ The total of 5 numbers is 24 × 5 = 120

The average of 3 numbers is 18
∴ The total of 3 numbers is 18 × 3 = 54

The sum of 8 numbers is = 120 + 54 = 174
∴ The average of 8 numbers is 174/8 = 21.75

Question: If tan(x - 30°) = 1, what is the value of sin(x + 15°)?

Solution:
Given that tan(x - 30°) = 1
⇒ (x - 30°) = 45°
⇒ x = 75°

Now,
sin(x + 15°)
= sin(75° + 15°)
= sin(90°)
= 1 

Question: What is the co-ordinate of the center of the circle (x - r)² + (y + 5)² = 100?

Solution: 
দেওয়া আছে, 
(x - r)² + (y + 5)² = 100
⇒ (x - r)² + {y - (- 5)}² = (10)²
অতএব, কেন্দ্র (a, b) = (r, - 5)

∴ বৃত্তের সমীকরণ হওয়ার শর্ত;
১। x² ও y² সহগ সমান হবে।
২। xy সমন্বিত কোন পদ থাকবে না।

Since it is a non - terminating and non - repeating decimal,
So it is an irrational number

Question: After a 10% reduction in the price of cooking oil, a family is able to purchase 4 liters more oil for Tk. 720. What was the original price per liter?

Solution:
Let
Original price of cooking oil = x Tk/liter.
Original quantity = 720/x​ litre

New price = 0.90x Tk/litre
New quantity = 720/0.90x = 720/(9x/10) = (720 × 10)/9x = 800/x​ liters

ATQ,
(800/x​) - (720/x) = 4
⇒ (800 - 720)/x = 4
⇒ 80/x = 4
⇒ x = 80/4
∴ x = 20

 ∴ Original price of cooking oil = 20 Tk/liter.

Question: x = 6, y = 4 and z = - 2, then x(y + z)/y(x + y + z) = ?

Solution:
Given that,
x =6, y = 4 and z = - 2

Then,
= x(y + z)/y(x + y + z)
= 6{4 + (- 2)}/{4(6 + 4 - 2)}
= 12/32
= 3/8

Question: A factory manufactures products in batches of 12, 18, and 30 units. What is the minimum number of units the factory needs to produce so that each batch can be formed exactly?

Solution:
To find the minimum number of units the factory needs to produce so that each batch size (12, 18, and 30) can be formed exactly, we need to find the least common multiple (LCM) of these batch sizes.

The prime factorization of each batch size is:
12 = 2 × 2 × 3 = 2² × 3¹

18 = 2 × 3 × 3 = 2¹ × 3²

30 = 2 × 3 × 5 = 2¹ × 3¹ × 5¹

Now,
The highest power of 2 is 2²
The highest power of 3 is 3²
The highest power of 5 is 5¹

So, the LCM of 12, 18, and 30 is:
= 2² × 3² × 5¹
= 4 × 9 × 5
= 180

∴ The minimum number of units the factory needs to produce is 180.

For option a, let's assume that A = 25, so option a can’t be entirely true
For option b, let's assume that A = 12, then option b is true
For option c, let's assume that A = 20, so option c is true
For option d, let's assume that A = 20, so option d is also true.
From the given condition of the options and question, option a can’t be inferred.

Question: How many two-digit numbers can be formed using the digits 3, 5, and 7 if repetition of digits is allowed?

Solution:
Number of possible two-digit numbers which can be formed by using the digits 3, 5 and 7 = 3 × 3.

∴ 9 possible two-digit numbers can be formed.

The 9 possible two-digit numbers are-
33, 35, 37, 53, 55, 57, 73, 75, 77 

Question: A train 300 meters long passes a pole in 20 seconds. How long will it take to pass a platform that is 500 meters long?

Solution:
Train's speed = Distance/Time
= 300/20 = 15m/s

Total distance to pass the platform,
= Length of train + Length of platform 
= 300 + 500
= 800m

∴ Required time = Distance/Speed
= 800/15
= 53.33 seconds

A opened 2 hours early to B
In 2 hours A can do 3 × 2 = 6 unit work
Remaining work = 24 - 6 = 18
A + B can do it in
= 18/5 hours
= 3 hours 36 minutes
∴ Tank will be full in 9 A.M. + 3 hours 36 minutes = 12.36 P.M.

Question: In a factory, 20 workers can make 20 toys in 10 days working 10 hours per day. Then, in how many days can 25 workers make 30 toys working 20 hours per day?

Solution:
Given:

Number of workers initially = 20
Number of days initially = 10
Hours per day initially = 10
Number of toys initially = 20
Number of workers later = 25
Hours per day later = 20
Number of toys later = 30
Number of days later = ?

∴ (Number of workers initially × Number of days initially × Hours per day initially × Number of toys later)
= (Number of workers later × Number of days later × Hours per day later × Number of toys initially)
⇒ 20 × 10 × 10 × 30 = 25 × (Number of days later) × 20 × 20
⇒ 20 × 10 = 200 → 200 × 10 = 2,000 → 2,000 × 30 = 60,000
⇒ 25 × 20 × 20 = 10,000 × (Number of days later)
⇒ 60,000 = 10,000 × (Number of days later)

∴ Number of days later = 60,000 ÷ 10,000 = 6 days

Question: The highest common factor (HCF) of x³ - 8, x⁴ - 16, and x³ - 2x² + 4x - 8 is:

Solution:
১ম রাশি = x³ - 8
= x³ - 2³
= (x - 2)(x² + 2x + 4)

২য় রাশি = x⁴ - 16
= (x²)² - 4²
= (x² - 4)(x² + 4)
= (x - 2)(x + 2)(x² + 4)

৩য় রাশি = x³ - 2x² + 4x - 8
= (x³ - 2x²) + (4x - 8)
= x²(x - 2) + 4(x - 2)
= (x² + 4)(x - 2)
= (x - 2)(x² + 4)

∴ নির্ণেয় গ.সা.গু (HCF) = (x - 2)

According to the question,
Chromium : Steel -
Type 1 - 2 : 11
Type 2 - 5 : 21
Now using alligation,

Ratio of quantity → 1 : 2

Question: A reservoir has two pipes, A and B. A can fill the reservoir 5 hours faster than B. If both together fill the reservoir in 6 hours, the reservoir will be filled by A alone in-

Solution: 
Let, A alone can fill the reservoir in x hours 
B can fill in x + 5 hours 

Both complete in 1 hour = (1/x) + (1/ x + 5)
= (2x + 5)/(x2 + 5x)

Now
1/{(2x + 5)/(x2 + 5x)} = 1/6
 (x2 + 5x)/(2x + 5) = 6 
⇒ x2 + 5x = 12x + 30 
⇒ x2 - 7x - 30 = 0
⇒ x2 - 10x + 3x - 30 = 0 
⇒ x (x - 10) + 3 (x - 10) = 0
⇒ (x - 10) (x + 3) = 0 
∴ x = 10 or, x = -3 , negative value not possible 

A alone can fill the reservoir in 10 hours

ধরি কর্মীর বেতন ছিল 100 টাকা
10% হ্রাসে তার বর্তমান বেতন হবে = (100 - 100 এর 10%) = [100 − {(100×10)/100}] =100 − 10 = 90 টাকা
আবার, 10% বৃদ্ধিতে তার বর্তমান বেতন হবে = (90 + 90 এর 10%) = [90 − {(90×10)/90}] = 90 + 9 = 99 টাকা
∴ তার মোটের উপর ক্ষতি হলো = 100 - 99 = 1 টাকা বা 1%

Let the numbers be a and b. Then, a + b = 12 and ab = 35.
(a + b)/ab = 12/35
arrow 1/b + 1/a = 12/35
Sum of reciprocals of given numbers = 12/35

Suppose they meet x hours after 7 a.m.
Distance covered by A in x hours = 20x km.
Distance covered by B in (x-1) hours = 25(x-1) km.
20x+25(x-1) = 110
45x = 135
So, x = 3.
So, they meet at 10 a.m.

Question: Mr. Rahim can complete a task in 12 days, while his daughter can complete the same task in 18 days. How many days will it take for them to finish the work if they work together?

Solution:
Given,
Mr. Rahim can do the work in 12 days
∴ in 1 day he can do = 1/12 part

His daughter can do the work in 18 days
∴ in 1 day she can do = 1/18 part

∴ in 1 day, working together they can do = (1/12 + 1/18) part
= (3 + 2)/36 part
= 5/36 part

Hence, together they complete 5/36 of the work in 1 day
∴ They can complete the whole work in = 36/5 days
= 7.2 days

∴ If they work together, the work will be completed in 7.2 days.

প্রশ্ন: প্রদত্ত 

সমাধান:

Question: Rita invested equal amounts at 5% and 7% simple interest for 4 years. If she received total interest of Tk. 1920, what was the amount invested in each scheme?

Solution:
Let the amount invested in each scheme = x taka
So, same amount at 5% and same amount at 7%

Interest at 5% for 4 years, 
I₁ = (x × 5 × 4)/100
= 20x/100
= x/5
= 0.20x

And, Interest at 7% for 4 years,
I₂ = (x × 7 × 4)/100
= 28x/100
= 0.28x

Total interest received = I₁ + I₂
0.20x + 0.28x = 1920
⇒ 0.48x = 1920
⇒ x = 1920/0.48
⇒ x = 1920 × (100/48)
⇒ x = 1920 × (25/12)
⇒ x = 160 × 25
∴ x = 4000

So, the amount invested in each scheme Tk. 4000

Question: If 3 men or 9 boys can finish a task in 36 days, how long will it take 9 men and 9 boys to finish the same task?

Solution:
এখানে, 3 জন পুরুষ = 9 জন বালক
∴ 1 জন পুরুষ = 9/3 = 3 জন বালক

এখন, 9 জন পুরুষ + 9 জন বালক
= (9 × 3) জন বালক + 9 জন বালক
= 27 জন বালক + 9 জন বালক
= 36 জন বালক

প্রশ্নমতে,
9 জন বালক কাজটি করে 36 দিনে
∴ 1 জন বালক কাজটি করে (36 × 9) দিনে
∴ 36 জন বালক কাজটি করে (36 × 9)/36 দিনে
= 9 দিনে

∴ কাজটি 9 দিনে শেষ হবে।

Question: An inlet pipe can fill a tank completely in 18 hours. In what time will the pipe fill 2/3 part of the tank?

সমাধান:
দেওয়া আছে,
ইনলেট পাইপটি সম্পূর্ণ ট্যাঙ্কটি পূর্ণ করতে সময় নেয় 18 ঘন্টা।

∴ ট্যাঙ্কটির 2/3 অংশ পূর্ণ করতে সময় লাগবে = (টোটাল সময় × অংশের পরিমাণ)
= (18 × 2/3) ঘন্টা
= (6 × 2) ঘন্টা
= 12 ঘন্টা।
∴ ট্যাঙ্কটির 2/3 অংশ পূর্ণ করতে 12 ঘন্টা সময় লাগবে।

Question: What is the angle between the hour and minute hands of a clock when it is 2 : 30 pm?

Solution:
2টা 30 মিনিট = 2 + (30/60) ঘন্টা
= 2 + 1/2 = 5/2 ঘন্টা

আমরা জানি,
ঘণ্টার কাঁটা 12 ঘণ্টায় 360° ঘোরে।
∴ 1 ঘণ্টায় ঘোরে = 360°/12 = 30°
∴ 5/2 ঘণ্টায় ঘোরে = (30° × 5)/2
= 150°/2 = 75°

আবার,
মিনিটের কাঁটা 60 মিনিটে 360° ঘোরে।
∴ 1 মিনিটে ঘোরে = 360°/60 = 6°
∴ 30 মিনিটে ঘোরে = 30 × 6° = 180°

∴ ঘড়ির কাঁটা দুটির মধ্যবর্তী কোণ = |180° - 75°| = 105°

Question: A pizza is divided into 36 slices. If Rahim takes 1/4 of the pizza and Karim takes 1/3 of the remaining slices, how many slices are still left?

সমাধান:
• মোট স্লাইসের সংখ্যা ছিল 36 টি।
• রহিম নেয় মোট স্লাইসের 1/4 অংশ।
⇒ রহিমের নেওয়া স্লাইস = 36 × (1/4) = 9টি।

∴ অবশিষ্ট স্লাইসের সংখ্যা = 36 - 9 = 27 টি।

• করিম অবশিষ্ট স্লাইসের 1/3 অংশ নেয়।
⇒ করিমের নেওয়া স্লাইস = 27 × (1/3) = 9 টি।

∴ সবশেষে অবশিষ্ট স্লাইসের সংখ্যা = 27 - 9 = 18 টি।

Here
b = √a
Or, we can write as b² = a
Now, c = a + b
or, c = b² + b
or, c = b(b + 1); now looking at the equation we need the value b as two consecutive integer then only the equation will satisfy.

Therefore, c is the product of two consecutive integers
I. 21 = 4 × 5 = 20 - NOT POSSIBLE
II. 30 = 5 × 6 = 30 - POSSIBLE
III. 45 = 6 × 7 = 42 - NOT POSSIBLE
IV. 331 = 18 ×19 = 342 - NOT POSSIBLE

Question: In a certain office, 1/3 of the workers are women, 1/2 of the women are married and 1/3 of the married women have children. If 3/4 of the men are married and 2/3 of the married men have children, what part of workers are without children?

Solution:
Given that,
Total women = 1/3
Married women = 1/2 of 1/3 = 1/6
Women who have a child = 1/3 of married women

∴ Women who has child = 1/3 of 1/6 = 1/18

And,
Total men = 2/3       (∵ 1/3 are women)
Married man = 3/4 of the total man
Married man = 3/4 of 2/3 = 1/2
Man who has a child = 2/3 of a married man

∴ Men who has child = 2/3 of 1/2 = 1/3

Now,
Men + Women (​​​​having child) = (1/18) + (1/3) = 7/18

∴ Part that don't have child = 1 - (7/18) = 11/18

∴ 11/18 workers don't have children.

1-minute work done by both the punctures = (1/9) + (1/6)
= (4 + 6)/36
= 10/36
= 5/18

So both punctures will make the type flat in = (18/5) min
= 3(3/5) min

21600=2⁵×3³×5²
To make it a perfect cube, it must be multiplied by (2 × 5), i.e., 10

Let the number of oranges in first basket be x,
Number of oranges in second basket = 672 - x
ATQ, x - x/5 = 672 - x + x/5
⇒ 4x/5 = 672 - 4x/5
⇒ 4x/5 + 4x/5 = 672
⇒ 8x/5 = 672
⇒ x = 672× (5/8)
⇒ x = 420
∴ Number of oranges in first basket = 420.