Bank Math

Let the required number of rounds be x
More radius, Less rounds (Indirect proportion)
∴ 20:14::70:x
⇔(20×x)=(14×70)
⇔ x= (14×70)/ 20
⇔ x=49

Question: A dishonest milkman professes to sell his milk at cost price but he mixes it with water and thereby gains 25%. The percentage of water in the mixture is-

Solution:
A dishonest milkman professes to sell his milk at cost price, but he mixes it with water and gains 25%

We know, 
Profit = SP - CP
Let milk bought (cost price) be 1 litres at Tk. 100 and profit = 25

∴ Selling price = 100 + 25 = Tk. 125

As the selling price of 1 litre was Tk. 100(same as cost)

Quantity sold = 125/100 = 5/4 = 1.25 litre

Hence, water added = 1.25 - 1 = 0.25 litres

∴ Percentage of water = (0.25/1.20) × 100 = 20%

The year 2004 is a leap year. It has 2 odd days.
∴ The day on 8th Feb, 2004 is 2 days before the day on 8th Feb, 2005.
Hence, this day is Sunday.

Question: The average of the ages of a man and his daughter is 34 years. If the respective ratio of their ages four years from now is 14 : 5. What is daughter's present age ?

Solution:
Average age of man and his daughter = 34 years
Their total age = (34 × 2) years = 68 years
Let man's age be x years. Then, daughter age = (68 - x) years
∴ (x + 4)/(68 - x + 4) = 14/5
⇒ 5(x + 4) = 14(72 - x)
⇒ 5x + 20 = 1008 - 14x
⇒ 19x = 988
⇒ x = 52

∴ Daughter's present age = (68 - 52) = 16 years.

Relative speed = (120 + 80) km/hr
(200 times; 5/18) m/s
(500/9) m/s

Let the length of other train x meter
Then, (x +270)/9 = 500/9
⇒ x + 270 = 500
⇒ x = 230

Answer : 230 m

Question: How many prime numbers are there between 70 and 80?

Solution:
যে সংখ্যা ১ এবং সেই সংখ্যা ছাড়া অন্য কোনো সংখ্যা দ্বারা বিভাজ্য নয়, তাকে মৌলিক সংখ্যা বলে।

৭০ থেকে ৮০ এর মধ্যে সংখ্যাগুলো হলো:
৭১, ৭২, ৭৩, ৭৪, ৭৫, ৭৬, ৭৭, ৭৮, ৭৯

এদের মধ্যে মৌলিক সংখ্যা:
৭১, ৭৩, ৭৯

অতএব, মোট মৌলিক সংখ্যা = ৩টি

Question: In a party there are 5 couples. Out of them 5 people are chosen at random. Find the probability that there are at the least two couples?

Solution:
Number of ways of (selecting at least two couples among five people selected) = ⁵C₂ × ⁶C₁
As remaining person can be any one among three couples left.
∴ Required probability = (⁵C₂ × ⁶C¹)/¹⁰C₅
= (10 × 6)/252
= 60/252
= 5/21

Question: If (2 + √x) > 2√x, which of the following must be true?

Solution:
2 + √x > 2√x
⇒ 2 > 2√x - √x
⇒ 2 > √x
⇒ 4 > x
∴ x < 4

Question: How many times is the area of the square constructed on a straight line greater than the area of the square constructed on one-third of that line?

Solution:
Let the length of the straight line be a units.
∴ Area of the square built on the whole line = a²

And, one-third of the line = a/3
∴ Area of the square built on one-third of the line = (a/3)² = a²/9

∴ Ratio = (square on whole line)/(square on one-third)
= a²/(a²/9)
= a² × (9/a²)
= 9

So the square on the whole line is 9 times the square on one-third of the line.

Question: 

Solution: 

Question: What is the H.C.F. of 9/20, 15/28, and 21/35?

Solution:
We know, H.C.F. of fractions = (H.C.F. of numerators) / (L.C.M. of denominators)
H.C.F of numerators: H.C.F of 9, 15 and 21
9 = 3 × 3 
15 = 3 × 5
21 = 3 × 7
H.C.F = 3

L.C.M of denominators: L.C.M of 20, 28 and 35
20 = 2² × 5
28 = 2² × 7
35 = 5 × 7
L.C.M = 2² × 5 × 7 = 140

∴ Required H.C.F. = 3/140

Question: The ratio of the present ages of two friends is 3 : 5. After 7 years, the ratio becomes 2 : 3. What will be the ratio of their ages after 10 years?

Solution:
Let
The present age of the friend₁ = 3x
The present age of the friend₂ = 5x

After 7 years, their ages will be,
⇒ (3x + 7)/(5x + 7) = 2/3
 ⇒ 10x + 14 = 9x + 21
 ⇒ 10x - 9x = 21 - 14
∴ x = 7

Present age of friend₁ = 3x = 21 years
Present age of friend₂ = 5x = 35 years

Now, after 10 years,
friend₁ age = 21 + 10 = 31 years
friend₂ age = 35 + 10 = 45 years

∴ The ratio of their ages after 10 years,
= 31 : 45

Let Nila's age be 5x years and
Shila's age be 6x years

((1/3)×5x):((1/2)×6x) = 5:9
⇒ 5x/(3×3x) = 5/9

Thus, Shila's age cannot be determined

Total students i.e. Section A + Section B = 80
Section A = 80 - Section B = 80 - B

∴ 80 x 35 = B x 30 + (80 - B) x 55
⇒ 2800 = 30B + 4400 - 55B
⇒ 1600 = 25 B
⇒ B = 64

∴ B = 64 = Number of students in Class V B.

Question: ABCD is a square and one of its sides AB is also a chord of the circle as shown in the figure. What is the area of the square?


Solution:
চিত্রানুসারে, O হলো বৃত্তের কেন্দ্র এবং OA ও OB হলো বৃত্তের ব্যাসার্ধ, যার দৈর্ঘ্য 3।
 AOB একটি সমকোণী ত্রিভুজ, যেখানে ∠AOB = 90° এবং অতিভুজ = AB

পিথাগোরাসের উপপাদ্য অনুসারে,
AB² = OA² + OB²
AB² = 3² + 3²
AB² = 9 + 9
AB² = 18

আমরা জানি, বর্গক্ষেত্রের ক্ষেত্রফল = বাহুর দৈর্ঘ্য^২ 
যেহেতু ABCD একটি বর্গ, তাই এর ক্ষেত্রফল হলো AB²
সুতরাং, বর্গটির ক্ষেত্রফল হলো 18

Interest for 1 year is the same whether it's simple interest or compound interest.
Now interest of third-year = 8575 - 7350 = 1225;
means principal for this interest is 7350 if compound interest is taken 7350 is the principal interest = 1225

if 100 is the principal interest = (1225/7350) × 100 = 50/3%
When a thing increases for two successive times the overall increase on initial amount = a + b + (a × b)/100

Therefore overall interest for two years = 50/3 + 50/3 + [(50/3) × (50/3)]/100 = 325/9 %
Therefore amount after 2 years = 100 + 325/9 = 1225/9

If 1225/9 is the amount principal =100
if 7350 is the amount principal =(900/1225) × 7350 =5400

So sum = 5400

Let the speed of the boat in still water=x km/hr
Speed of the current = 2 km/hr
Then, speed downstream = (x + 2) km/hr
speed upstream = (x - 2) km/hr
Total time taken to travel 10 km upstream and back = 55 minutes
= (55/60) hr
= 11/12 hr.
According to question,
10/(x - 2) + 10/(x + 2) = 11/12
⇒ 120(x+2)+120(x−2) = 11(x² - 4)
⇒ 240x = 11x² - 44
⇒ 11x² - 240x - 44 = 0
⇒ 11x(x -22) + 2(x - 22) = 0
⇒ (x -22) (11x + 2) = 0
Since x cannot be negative.
So, x = 22 km/hr.
Hence, the Speed of the motorboat is 22 km/hr.

Question: Two numbers are in the ratio 4 : 5 and their greatest common divisor (GCD) is 14. Find their least common multiple (LCM).

Solution:
Let the two numbers are,
First number = 4k
Second number = 5k
Since the numbers are in the ratio 4 : 5, and 4 and 5 are co-prime gcd(4,5) = 1,
the HCF of the two numbers is exactly k.

Given that, HCF = 14
Therefore, k = 14

So the actual numbers are:
First number = 4 × 14 = 56
Second number = 5 × 14 = 70

We know,
HCF × LCM = Product of the two numbers
⇒ 14 × LCM = 56 × 70
⇒ LCM = (56 × 70)/14
∴ LCM = 4 × 70 = 280

So the least common multiple (LCM) of the two numbers is 280.

Since, no order to the committee is mentioned, a combination instead of a permutation is used.
Let's sort out what we have and what we want.
Have: 5 women, 6 men.
Want: 3 women AND 1 man.
The word AND means multiply.
Woman and Men

⁵C₃ × ⁶C₁
= 5!/(2! × 3!) × 6!/(5! × 1!)
= (5 × 2) × 6
= 60.

Question: From a point P on a level ground, the angle of elevation of the top of a tower is 30°. If the tower is 100 m high, the distance of point P from the foot of the tower is-

Solution:

Given that,
Hight of tower, AB = 100 m
Angle of elevation from point P, ∠APB = 30°

We know,
tanθ = opposite/adjacent ​= AB/PA
⇒ tan30° = 100/PA
⇒ 1/√3 = 100/PA
∴ PA = 100√3 m

Thus, the distance from point P to the foot of the tower is 100√3 m.

Let, Asad have X taka
ATQ,
X/10 - X/12 = 3
⇒ 6X - 5X = 60×3
⇒ X = 180

আমরা দেখতে পাচ্ছি যে চারটি পরস্পর সমান্তরাল রেখাকে তিনটি পরস্পর সমান্তরাল রেখা ছেদ করলে মোট আয়তক্ষেত্রের সংখ্যা হবেঃ
6 + 5 + 4 + 3 + 2 + 2 + 1 = 18