Bank Math

Question: What is the percentage increase in the area of a rectangle, if each of its side is increased by 25%?

Solution:
মনে করি,
আয়তক্ষেত্রের দৈর্ঘ্য x একক
আয়তক্ষেত্রের প্রস্থ y একক
∴ আয়তক্ষেত্রের ক্ষেত্রফল = xy বর্গ একক

25% বৃদ্ধিতে,
আয়তক্ষেত্রের  নতুন দৈর্ঘ্য = {x + x এর 25%} একক
= {x + x এর (25/100)} একক
= {x + x/4} একক
= 5x/4 একক

25% বৃদ্ধিতে,
আয়তক্ষেত্রের  নতুন প্রস্থ = {y + y এর 25%} একক
= {y + y এর (25/100)} একক
= 5y/4 একক

আয়তক্ষেত্রের নতুন ক্ষেত্রফল = (5x/4) × (5y/4) বর্গ একক
= (25xy/16) বর্গ একক

ক্ষেত্রফল বৃদ্ধি পায় = {(25xy/16) - xy} বর্গ একক
= (25xy - 16xy)/16 বর্গ একক
= 9xy/16 বর্গ একক

ক্ষেত্রফল xy বর্গ এককে বৃদ্ধি পায় 9xy/16 বর্গ একক
ক্ষেত্রফল 1 বর্গ এককে বৃদ্ধি পায় (9xy/16) × (1/xy) বর্গ একক
ক্ষেত্রফল 100 বর্গ এককে বৃদ্ধি পায় (9xy/16) × (1/xy) × 100 বর্গ একক
= 56.25 বর্গ একক

Question: Find a number that is divisible by 7, but leaves remainder 5 when divided by 8, 12, and 20.

Solution:
Let the number be N.
It satisfies:
N ≡ 0 (mod 7)

If N leaves remainder 5 when divided by 8, 12, and 20, then N - 5 is divisible by all three numbers:
N - 5 ≡ 0 (mod 8, 12, 20)

Find LCM of 8, 12, 20:
8 = 2³
12 = 2² × 3
20 = 2² × 5
LCM = 2³ × 3 × 5 = 120

N must also be divisible by 7:
120k + 5 ≡ 0 (mod 7)  
⇒ 120k ≡ -5 ≡ 2 (mod 7)
(since -5 ≡ 2 mod 7)

Smallest positive k = 2 →
N = 120 × 2 + 5 = 240 + 5 = 245

Question: If a wire of 440 metres length is moulded in the form of a circle and a square turn by turn, find the ratio of the area of the circle to that of the square.

Solution: 
Given that, 
Length of wire = 440 m

Circumference of circle = 440 m
2πr = 440 
r = 440/{(22/7) × 2} = (440 × 7)/44 = 10 × 7 = 70
∴ r = 70

∴ Area of circle = πr² = (22/7) × 70² = (22/7) × 70 × 70 = 15400 m²

And,
Perimeter of square = 440 m
4a = 440
⇒ a = 440/4 = 110 
∴ a = 110

∴ Area of square = a² = 110² = 12100 m²

∴ Required ratio of Area of circle : Area of square = 15400 : 12100 = 154 : 121 = 14 : 11

So the ratio of the area of the circle to that of the square is 14 : 11.

Consider the diagram is shown above where QR represents the tree and PQ represents its shadow.
We have, QR = PQ
Let, ∠QPR = θ
tanθ = QR/PQ
= QR/QR [since QR = PQ]
= 1
= tan 45°
⇒ θ = 45°
i.e., the required angle of elevation = 45°

Probability = What we want/Total
Or = Add
And = Multiply

Both white OR both blue
Total shirts = 15
There are 2 white and 3 blue shirts

Probability for 2 white shirts = (2/15) × (1/14) = 1/105
Probability for 2 blue shirts = (3/15) × (2/14) = 1/35

∴ Total probability = (1/105) + (1/35) = 4/105.

Quantity of blood donated in 2 years
= (350 × 3) ml
= 1050 ml
= 1.05 litres
∴ Quantity of blood donated in 6 years
=( 1.05/2 ×6)
= 3.15 litres

Question: Two boats are travelling towards each other at speeds of 46 km/h and 62 km/h, respectively. What is the distance between the two boats half a second before they collide?

Solution: 
Relative speed = (46 + 62) km/h
= 108 km/h
= [108 × (5/18)] m/s
= 30 m/s

We know,
Distance = Relative speed × Time
= [30 × (1/2)] m
= 15 m 

∴ Distance between the two boats half a second before collision = 15 meters 

Question: If 3^{x + 3} + 7 = 250, then x is equal to?

Solution:
Given that, 
3^{x + 3} + 7 = 250
⇒ 3^{x + 3} = 250 - 7
⇒ 3^{x + 3} = 243
⇒ 3^{x + 3} = 3⁵
⇒ x + 3 = 5
⇒ x = 5 - 3
∴ x = 2

So the value of x is 2

Question: In the following arithmetic sequence: 5, 12, 19, 26, ...... what is the place of the term with a value of 145?

Solution:
Given,
n th term = 145
First term (a) = 5
Common difference (d) = 12 - 5 = 7
We know,
n th term = a + (n - 1)d
⇒ 145 = 5 + (n - 1) × 7
⇒ 145 = 5 + 7n - 7
⇒ 145 = 7n - 2
⇒ 145 + 2 = 7n
⇒ 147 = 7n
⇒ n = 147/7
∴ n = 21

∴ 21 th term holds the value of 145.

Question: A man completes a certain journey by a car. If he covered 30% of the distance at the speed of 20kmph. 60% of the distance at 40km/h and the remaining of the distance at 10 kmph, his average speed is-

Solution:

Question: What should come in place of both the question marks in the equation ?

Solution:

ধরি,একটি কলমের মূল্য = 100 টাকা
তাহলে 2 টি কলমের মূল্য = 100 - 100×20% = 100 - 20 = 80 টাকা
সুতরাং 1 টি কলমের মূল্য = 80/2 = 40 টাকা
সুতরাং কলমঃপেন্সিল = 100:40 = 5:2.

Question: In a class, if 5 students are seated on each bench, 5 benches remain vacant. But if 4 students are seated on each bench, 8 students have to stand. How many students are there in the class?

Solution:
Let the number of benches = b

1^st condition,  
5 students are seated per bench then 5 benches remain vacant  
∴ Students = 5 × (b - 5)

2^nd condition, 
4 students are seated per bench then 8 students are left standing  
∴ Students = 4 × b + 8

ATQ,
5(b - 5) = 4b + 8
⇒ 5b - 25 = 4b + 8  
⇒ 5b - 4b = 8 + 25  
∴ b = 33

Now, total students = 5 × (33 - 5)  
= 5 × 28  
= 140

∴ There are 140 students in the class.

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

Solution:
To find the minimum number of units the bakery 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 × 3
18 = 2 × 3 × 3 = 2 × 3²
30 = 2 × 3 × 5 = 2 × 3 × 5

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 × 5
= 4 × 9 × 5 = 180

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

Question: There are 8 multiple-choice questions in an examination. Each question has 4 options. In how many different ways can these questions be answered?

Solution:
For each multiple-choice question, there are 4 possible answers (options A, B, C, or D).

Since there are 8 questions, and each question can be answered independently:

Total number of ways = 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4
= 4⁸ ways
= 65536 ways

Question: A geometric series has its first term as 1 divided by square root of 2, and its common ratio is √2. Which term in the sequence is 32√2?

Solution:
First term, a = 1/√2
Common ratio, r = √2

Let, the n-th term be = ar^{n - 1} = 32√2
⇒ (1/√2) (√2)^{n - 1} = 32√2
⇒ (√2)^{n - 1} = 32√2 × √2
⇒ (√2)^{n - 1} = 32 × 2
⇒ (√2)^{n - 1} = 64
⇒ (√2)^{n - 1} = (√2)¹²
⇒ n - 1 = 12
∴ n = 13

∴ The 13th term is 32√2.

We can write three-quarters of a kilometre as 750 metres, 11(1/4) minutes as 675 seconds and 7(1/2) minutes as 450 seconds.

Rate upstream = (750/675) m/sec = 10/9 m/sec
Rate downstream = (750/450) m/sec = 5/3 m/sec
Rate in still water = (1/2) (10/9 + 5/3) m/sec
= 25/18 m/sec
= (25/18 × 18/5) km/hr
= 5km/hr

Since he make 8% profit so his purchase = 100 + 8 = 108
He make 8% more profit during sales so his sales = (108/100) × 108 = 116.64
so profit = 116.64 - 100
= 16.64
Short Cut (with increase decrease short cut)
X + Y + (xy/100)
= 8 + 8 + (8×8/100) ( since he his making profit so count is as a increase)
= 16.64

let x = 5 and y = 3

Then,
(8x - 5y) : (8x + 5y)
= {(8 × 5) - (5 ×3)} : {(8 × 5} + (5 × 3)}
= (40 - 15) : (40 + 15)
= 25 : 55
= 5 : 11.

Question: P, Q, and R share a total of Tk. 2100 such that P receives twice as much as Q, and Q receives three times as much as R. How much money does Q get?

Solution:
Given,
P = 2Q
Q = 3R

∴ P = 2 × 3R = 6R

So, the ratio of P : Q : R = 6 : 3 : 1

∴ Q receives = (2100 × 3/10) Tk
= 630 Tk

∴ Q receives Tk. 630.

Question: Sumon and Noyon can complete a task together in 6 days. Noyon alone can do it in 9 days. How many days will Sumon take to complete the task alone?

Solution:
ধরা যাক, সুমন একা কাজটি করতে x দিন সময় নেয়।
∴ সুমনের 1 দিনের কাজ = 1/x অংশ
∴ নয়নের 1 দিনের কাজ = 1/9 অংশ (যেহেতু সে একা ৯ দিনে করে)

তারা দুজনে মিলে 6 দিনে কাজটি শেষ করে।
∴ তাদের একত্রে 1 দিনের কাজ = 1/6 অংশ

শর্তমতে,
(1/x) + (1/9) = 1/6
⇒ 1/x = (1/6) - (1/9)
⇒ 1/x = (3 - 2)/18
⇒ 1/x = 1/18
⇒ x = 18

∴ সুমন একা কাজটি করতে 18 দিন সময় নেবে।

In still water, we know that
speed of boat (x) = (1/2) × (Downstream speed + upstream speed)
With the given parameters like S_D = 40 km/hr & S_U = 22 km/hr, on substituting these values in above equation,
we obtain,
x = (1/2) × [S_D + S_U]
⇒ x = (1/2) × (40 + 22)
⇒ x = 31 km/hr.

প্রশ্ন: If log_x(81/16) = - 2, then x = ?

সমাধান:
দেওয়া আছে,
log_x(81/16) = - 2
⇒  x^{- 2} = 81/16
⇒  x^{- 2} = (9/4)²
⇒ x^-2 = (9/4)²
⇒  x^{- 2} = 1/(4/9)²
⇒  x^{- 2} = (4/9)^{- 2}
∴ x = 4/9

Question: Solve the inequality: 5(x - 3) + 7 < 3(2x - 1).

Solution:
Given inequality,
5(x - 3) + 7 < 3(2x - 1)
⇒ 5x - 15 + 7 < 6x - 3
⇒ 5x - 8 < 6x - 3
⇒ 5x - 6x < - 3 + 8
⇒ - x < 5
∴ x > - 5 

Question: In a 120-liter mixture of milk and water, the ratio of milk to water is 3 : 2. How many liters of water must be added to make the ratio become 1 : 3?

Solution:
Total mixture = 120 litres
Given ratio (milk : water) = 3 : 2

Milk = 120 × (3/5) = 72 liters
Water = 120 - 72 = 48 liters

To make the ratio 1 : 3, x liters of water need to be added.
milk : water = 72 : (48 + x)

So,
72/(48 + x) = 1/3

Cross-multiplying,
3 × 72 = 48 + x
⇒ 216 = 48 + x
⇒ x = 216 - 48
∴ x = 168

Quantity of water to be added = 168 liters.

Question: A barrack has enough food for 200 soldiers or 400 sailors. If 120 sailors have taken the food, how many soldiers will be able to eat with the remaining food?

Solution:
এখানে,
400 জন নাবিকের খাবার = 200 জন সৈন্যের খাবার।

মোট নাবিক যাদের জন্য খাবার ছিল = 400 জন।
খাবার গ্রহণ করেছে = 120 জন।
অবশিষ্ট খাবার = (400 - 120) জন নাবিকের খাবার
= 280 জন নাবিকের খাবার।

এখন,
400 জন নাবিকের খাবার = 200 জন সৈন্যের খাবার।
∴ 1 জন নাবিকের খাবার = (200/400) জন সৈন্যের খাবার।
∴ 280 জন নাবিকের খাবার = (200 × 280)/400
= (1/2) × 280
= 140 জন সৈন্যের খাবার।

সুতরাং, অবশিষ্ট খাবার দিয়ে 140 জন সৈন্যকে দেওয়া যাবে।

According to question,
√0.00005746
⇒ √5746100000000
⇒ 75.8/10000
⇒ 0.00758

Question: A 286 cm long copper strip is bent into a round wheel. What is the wheel’s diameter?

Solution:
ধরি,
গোলাকার চাকার ব্যাসার্ধ = r
ব্যাস = 2r
পরিধি = 2πr

প্রশ্নমতে, 
চাকার পরিধি = তামার তারের দৈর্ঘ্য
2πr = 286
⇒ 2r = 286/π
⇒ 2r = 286/(22/7)
⇒ 2r = (286 × 7)/22
⇒ 2r = 13 × 7
⇒ 2r = 91

∴ গোলাকার চাকার ব্যাস = 91 সে.মি.

Question: P runs three times faster than Q, and Q runs twice as fast as R. If R covers a certain distance in 120 minutes, how long will it take P to cover the same distance?

Solution:
Let,
Speed of R = x
Then speed of Q = 2x
Speed of P = 2x × 3 = 6x

Ratio of speeds P : Q : R = 6x : 2x : x = 6 : 2 : 1

Since time is inversely proportional to speed,
Ratio of time taken P : Q : R = 1/6 : 1/2 : 1 = 1 : 3 : 6
Given that,
R covers a certain distance in 120 minutes.
So,
6 units = 120 minutes
1 unit = (120 ÷ 6) minutes = 20 minutes

∴ P will cover the same distance in 20 minutes.

Question: Two positive numbers are in the ratio 3 : 2. The product of their HCF and LCM is 3456. Find the sum of both the numbers.

Solution:
Let two numbers are 3a and 2a.

We know,
HCF × LCM = 1st no. × 2nd no.
⇒ 3456 = 3a × 2a
⇒ 3456 = 6a²
⇒ a² = 576
⇒ a = 24 

∴ Sum of both the numbers = 3a + 2a = 5a = 5 × 24 = 120

Question: A can complete a work in 20 days, while B can complete the same work in 30 days. If both A and B work together, in how many days will they complete the entire work?

Solution:
A's 1 day work = 1/20
B's 1 day work = 1/30
Together 1 day work = 1/20 + 1/30
= (3 + 2)/60 = 5/60 = 1/12

∴ Total time = 1/Combined work rate
= 1/(1/12) days
= 12 days

Given,Principal,P = tk. 1000
Compound rate,R = 20%per annum
= 20/2
= 10% half-yearly
Time = 18 month
= 3 half - years

Amount,A = {P × [1 + (R/100)]ⁿ}
= {1000 × [1 + (10/100)]³}
= (1000 × 11 × 11 × 11) ÷ (10 × 10 × 10)
A = tk.1331
Hence, compound interest = tk.1331

Question: If P = 216^{- 1/3} + 243^{- 2/5} + 256^{- 1/4}, then which one of the following is an integer?

Solution:
P = 216^{- 1/3} + 243^{- 2/5} + 256^{- 1/4}
= (6³)^{- 1/3} + (3⁵)^{- 2/5} + (4⁴)^{- 1/4
=} 6^{3(- 1/3)} + 3^{5(- 2/5)} + 4^{4(- 1/4)}
= 6^{- 1}+ 3^{- 2}+ 4^{- 1}
= (1/6)+ (1/9) + (1/4)
= (6 + 4 + 9)/36
∴ P = 19/36

Now,
Option (A): P/19 = (19/36)/19 = 1/36, not an integer. Reject.
Option (B): P/36 = (19/36)/36 = 19/36², not an integer. Reject.
Option (C): 36/P = 36/(19/36) = 36²/19, not an integer. Reject.
Option (D): 19/P = 19/(19/36) = 36, an integer. Correct.
Option (E): P = 19/36, not an integer. Reject.