Bank Math

Question: The sum of ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of second youngest
one?

Solution:
মনে করি,
সবচেয়ে ছোট শিশুটির বয়স x বছর

যেহেতু প্রত্যেক শিশুর জন্মের ব্যবধান 3 বছর

তাই 5 জন শিশুর বয়স হবে:
প্রথম অর্থাৎ সবচেয়ে ছোট শিশুটির বয়স  x বছর
দ্বিতীয় অর্থাৎ দ্বিতীয় ছোট শিশুটির বয়স (x + 3) বছর
তৃতীয় শিশুটির বয়স (x + 3 + 3) বা (x + 6) বছর
চতুর্থ শিশুটির বয়স (x + 6 + 3) বা (x + 9) বছর
পঞ্চম শিশুটির বয়স (x + 9 + 3) বা (x + 12) বছর

প্রশ্নমতে,
x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50
⇒ 5x + 30 = 50
⇒ 5x = 50 - 30
⇒ 5x = 20
⇒ x = 20/5
∴ x = 4 

∴ দ্বিতীয় কনিষ্ঠ অর্থাৎ দ্বিতীয় ছোট শিশুটির বয়স = (4 + 3) বছর
= 7 বছর

Question: If 75% of the students in a school are boys and the number of girls is 520, the number of boys is:

Solution:
No. of boys = 75%
No. Girls = 25% = 520

Now 
25% = 520
1% = 520/25
75% = (520 × 75)/25
= 1560

According to the question,
10% of 20 = 2 mangoes were damaged
∴ বিক্রি করলো 20 - 2
=18
= 1(1/2) dozen.
বিক্রয় মূল্য = 54 × 1(1/2)
= 81
∴ Profit = 81 - 55
= 26.

প্রথম স্থানে 9 স্থির রেখে তৃতীয় স্থানে 3 নির্দিষ্ট করে সাজানো যায় ⁸P₃ উপায়ে
প্রথম স্থানে 9 স্থির রেখে তৃতীয় স্থানে 6 নির্দিষ্ট করে সাজানো যায় ⁸P₃ উপায়ে
∴ মোট বিন্যাস সংখ্যা =  ⁸P₃ +  ⁸P₃ 
^{= (8×7×6×5! / 5!) + (8×7×6×5! / 5!)
= 336 + 336
= 672}

ধরি, বৃত্তের ব্যাসার্ধ r cm
এর ক্ষেত্রফল πr² cm²

প্রশ্নমতে, 2πr = 16π
r = 16/2 =8

ক্ষেত্রফল = πr²cm²
π × (8)² cm²
=64π cm²

Question: In how many different ways can the letters of the word 'ORATION' be arranged so that the vowels always come together?

Solution:
The word 'ORATION' contains 7 letters, where O is repeated twice.

The vowels are O, A, I, O.

When the vowels are always together, they can be supposed to form one letter.

Then, we have to arrange the letters RTN (OAIO).

Now, 4 letters can be arranged in 4! = 24 ways.

The vowels (O, A, I, O) can be arranged among themselves in = 4!/2! = 12 ways.

∴ Required number of ways = (24 × 12) = 288.

Question: A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time as the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is-

Solution:
Let the time taken by the first pipe = x hours

Then,
The second pipe fills the tank 5 hours faster than the first pipe
∴ Time = x - 5 hours
The third pipe fills the tank 4 hours faster than the second pipe
∴ Time = (x - 5) - 4 = x - 9 hours

ATQ,
(1/x) + {1/(x - 5)} = 1/(x - 9)
⇒ (x - 5 + x)/{x(x - 5)} = 1/(x - 9)
⇒ (2x - 5)(x - 9) = x(x - 5)
⇒ 2x² - 23x + 45 = x² - 5x 
⇒ x² - 18x + 45 = 0
⇒ x² - 15x - 3x + 45 = 0
⇒ (x - 3)(x - 15) = 0
∴ x = 3 or x = 15
Since time cannot be less than 9 hours
Hence, x = 15

Question: When a person weighing 60 kg leaves a group of 50 people, the average weight of the remaining people increases by 0.3 kg. What is the new average weight of the remaining 49 people?

Solution:
let,
The average weight of 49 people is x kg
Total weight of 49 people = 49x
Total weight of 50 people = 49x + 60

ATQ,
50(x - 0.3) = (49x + 60)
⇒ 50x - 15 = 49x + 60
⇒ 50x - 49x = 60 + 15
∴ x = 75

∴ the new average weight of the remaining 49 people is 75 kg.

Question: How many years will Tk. 1500 need at 4% simple interest to earn the same interest as Tk. 2000 earns in 3 years at 5% simple interest?

Solution:
Interest from Tk. 2000 in 3 years at 5%:
SI = (P × R × T)/100
= (2000 × 5 × 3)/100
= 300 Tk.

Now,
Tk. 1500 to Produce the Same Interest (Tk. 300) at 4%:
Principal, P = Tk. 1500
Interest, SI = Tk. 300
Interest Rate, R = 4%
​ Time, T = ? years

SI = PRT/100
⇒ T = (SI × 100)/(P × R)
⇒ T = (300 × 100)/(1500 × 4)
⇒ T = 30000/6000
∴ T = 5 years

Area of square = d²/2 (d for diagonal)
so, area of bigger square = x²/2
so area of smaller square = x²/4 (half of the bigger one)
one side of smaller square = √(x²/4)
= x/2
diagonal of smaller square d² = (x/2)² + (x/2)²
d = root 2 × x/2.

Question: If (a/b)^{3x - 4} = (b/a)^{x + 2}, then what is the value of x?

Solution:
Given, (a/b)^{3x - 4} = (b/a)^{x + 2}
⇒ (a/b)^{3x - 4} = (a/b)^{- (x + 2)}
⇒ (a/b)^{3x - 4} = (a/b)^{- x - 2}
⇒ 3x - 4 = - x - 2
⇒ 3x + x = - 2 + 4
⇒ 4x = 2
∴ x = 1/2

ⁿC_r= n!/(r!) (n – r)!
3 consonants out of 5 can be selected in ⁵C₃ ways
2 vowels out of 4 can be selected in ⁴C₂ ways
In ⁵C₃ and ⁴C₂ ways the consonants and vowels can be selected.
Remember ⁵C₃ and ⁴C₂, hence multiply the terms ⁵C₃ × ⁴C₂
⁵C₃ × ⁴C₂ = {5!/3!(5 - 3)!} × {4!/2!(4 - 2)!}
= {(5 × 4 × 3 × 2 × 1)/(3 × 2 × 2)} × (4 × 3 × 2 × 1)/(2 × 2)
=(5 × 2) × (3 × 2)
60 groups can be made each having 3 consonants and 2 vowels
Each word contains 3 consonants and 2 vowels i.e 5 letters.

If N different objects are to be arranged, then they can be arranged in N! ways.
5 letters can be arranged in 5! Ways.
5! = 5 x 4 x 3 x 2 x 1 = 120
Required number of words = (120 x 60) = 7200

Team won 40 games out of 60 and the remaining games were 32.
Total games = 60 + 32
= 92
75% of 92 = 69
Team has to win 69 games in total.
Team has already won 40.
∴ Games to win = 69 - 40
= 29

Question: What is the slope of a line parallel to the line whose equation is 2x + 5y = 10?

Solution:
প্রদত্ত সরলরেখার সমীকরণটি হলো: 2x + 5y = 10
কোনো সরলরেখার ঢাল নির্ণয়ের জন্য সেটিকে y = mx + c আকারে আনতে হয়, যেখানে m হলো ঢাল।
2x + 5y = 10
⇒ 5y = - 2x + 10
⇒ y = (- 2/5)x + 2
অতএব, প্রদত্ত রেখার ঢাল (m₁) = - 2/5

আমরা জানি, দুটি সমান্তরাল রেখার ঢাল সমান হয় (m₁ = m₂)। 
∴ সমান্তরাল রেখাটির ঢাল, m₂= - 2/5

Let the required number of days be x
7 men = 10 women
(14 men and 20 women) = (20 + 20) women = 40 women
More length, More days (Direct proportion)
More women, Less days (Indirect proportion)
Length 100:600 & Women 40:10} :: 10:x
∴ 100×40×x = 600×10×10
⇔ x = (600×10×10)(100×40)
⇔ x = 15

Question: Through what angle does the minute hand of a clock turn in 15 minutes?

Solution:
We know,
Angle traced by the minute hand in 60 minutes = 360°

∴ Angle of the minute hand in 15 minutes,
= (360 × 15)/60
= 90°

Let X be the number which is added to 24
85% of X = 0.85X
Now,
24 + 0.85X = X
⇒ 0.15X = 24
∴ X = 24/0.15 = 160

Question: Which fraction is greater than 2/5 and less than 8/9?

Solution: 
2/5 = 0.40
8/9 = 0.88

Now,
4/9 = 0.44;  greater than 0.40, less than 0.88
3/4 = 0.75 → greater than 0.40, less than 0.88
5/12 = 0.41;  greater than 0.40, less than 0.88

All the given fractions lie between 2/5 and 8/9.

So correct answer: d) All of them 

Question: What should be the value of "Q" so that the expression (49 - 28x + Qx²) becomes a perfect square?

Solution:
(49 - 28x + Qx²)
= (7)² - 2 × 7 × 2x + (2x)² + Qx² - (2x)²
= (7 - 2x)² + Qx² - 4x²

The expression becomes a perfect square if,
Qx² - 4x² = 0
⇒ Qx² = 4x² 
∴ Q = 4

Thus, when Q = 4, the expression is (7 - 2x)², which is a perfect square.

Space is vacant for (20 - 15) = 5 buses
20 buses = 32 cars
1 bus = (32/20) cars
5 buses = (32/20) × 5
= 8 buses

Given,
x = 1/2 as the root of equation x² - mx - 5/4 = 0.
(1/2)² – m(1/2) – 5/4 = 0
1/4 - m/2 - 5/4 = 0
m = -2

Question: Two numbers P and Q are such that the sum of 10% of P and 15% of Q is three-fourths of the sum of 20% of P and 18% of Q. Find the ratio of P : Q.

Solution:
10% of P + 15% of Q = 3/4 × (20% of P + 18% of Q)
⇒ 10P/100 + 15Q/100 = 3/4 × (20P/100 + 18Q/100)
⇒ P/10 + 3Q/20 = 3/4 × (P/5 + 9Q/50)
⇒ 2P/20 + 3Q/20 = 3/4 × (10P + 9Q)/50
⇒ (2P + 3Q)/20 = (30P + 27Q)/200
⇒ 10(2P + 3Q) = 30P + 27Q
⇒ 20P + 30Q = 30P + 27Q
⇒ 30Q - 27Q = 30P - 20P
⇒ 3Q = 10P
⇒ P/Q = 3/10
∴ P : Q = 3 : 10

Question: Two dice are tossed at once. What is the probability that their combined total is no less than 11?

Solution:
If two dice are rolled together total possible outcomes = 6 × 6 = 36

But we want sum ≥ 11

Sum of the pairs of dice , 
Sum =  11 where (5, 6), (6, 5) here 2 outcomes
Sum = 12 (6, 6) here 1 outcomes

∴ Total outcomes = 2 + 1 = 3

The probability = 3/36
= 1/12

Question: What will come at the place of question mark?
3, 10, 24, 45, 73, ?

Solution:
পার্থক্যগুলো বের করি:
10 - 3 = 7
24 - 10 = 14
45 - 24 = 21
73 - 45 = 28

প্রতিবার পার্থক্য 7 করে বৃদ্ধি পাচ্ছে।
∴ পরবর্তী পার্থক্য হবে = 28 + 7 = 35
∴ পরবর্তী সংখ্যা = 73 + 35 = 108

Question: In an examination A got 25% marks more than B, B got 10% less than C and C got 25% more than D. If D got 320 marks out of 500, the marks obtained by A were :

Solution:
Marks obtained by D = 320

Marks obtained by C = 320 × (125/100)
= 400

Marks obtained by B
 = 400 × (100 - 10)/100
= 360

Marks obtained by A
= 360 × (125/100)
= 450

∴ Hence, required marks obtained by A = 450

We know that,
Dividend = [(Divisor × Quotient)] + Remainder
It is given that the remainder is 3 when a number (dividend) is divided 5(divisor).
Dividend and quotient are unknown,

Hence assume dividend as X and quotient as Y.
X = (5Y) + 3
The square of this number is divided by 5, therefore
X² = (5Y + 3)²
X² = (25Y²+ 30Y + 9)

On dividing this equation,
we get,
= 25Y² + 30Y + 9
= 25Y² + 30Y + 5 + 4
= 5(5Y²+ 6Y + 1) + 4

The remainder is 4, because 9 is not exactly divisible by 5 and we get 4 as remainder.

Question: P, Q, and R opened a shop together. P invests four times as much as Q, and Q invests one-fifth as much as R. After one year, the total profit is Tk 18,000. What is Q's share of the profit?

Solution:
Let R's investment = x taka
Then Q's investment = 1/5 of x = x/5
And P's investment = 4 × (x/5) = 4x/5

So the ratio of investments (or profit) is:
P : Q : R
= 4x/5 : x/5 : x
= 4 : 1 : 5

Total ratio = 4 + 1 + 5 = 10

Total profit = Tk 18,000

∴ Q's share = (1/10) × 18000 = Tk 1800

Question: A and B are two fixed points 10 cm apart and C is a point on AB such that AC is 6cm. if the length of AC is increased by 10%, by what percentage is CB decreased?

Solution:
দেওয়া আছে 
AB = 10 cm
AC = 6 cm
CB = (10 - 6) = 4

AC 10% বৃদ্ধিতে = 6 + 6 এর 10%
= 6 + 6 এর 10/100 
= 6 + .6
= 6.6

CB এর নতুন দৈর্ঘ্য = 4 - . 6
= 3.4

CB এর দৈর্ঘ্য হ্রাস পায় = 4 - 3.4
= .6 

শতকরা হ্রাস পায় = (.6/4) × 100%
= 15%

Question: Taposh and Sumon can finish a work together in 10 hours. If Taposh takes 4 times as long as Sumon to finish the job alone, how long will Sumon take to finish the job alone?

Solution:
ধরি, সুমন একা কাজটি শেষ করতে সময় নেয় = x ঘণ্টা
তাপস একা কাজটি শেষ করতে সময় নেয় = 4x ঘণ্টা

সুমনের 1 ঘণ্টার কাজের পরিমাণ = 1/x অংশ
তাপসের 1 ঘণ্টার কাজের পরিমাণ = 1/(4x) অংশ

তারা একসাথে 1 ঘণ্টায় কাজ করতে পারে = (1/x) + 1/(4x)
= (4 + 1) / (4x)
= 5 / (4x) অংশ

একসাথে কাজটি শেষ করতে সময় লাগে = 10 ঘণ্টা
অর্থাৎ, 10 ঘণ্টায় তারা পুরো 1টি কাজ শেষ করে

শর্তমতে,
(5/4x) × 10 = 1
⇒ 50/4x = 1
⇒ 4x = 50
⇒ x = 50 / 4
⇒ x = 12.5

∴ সুমনের কাজটি একা শেষ করতে 12.5 ঘণ্টা সময় লাগবে।

Question: The difference between compound interest and simple interest on a certain sum for 2 years at 12% per annum is 36 Taka. What is the principal amount?

Solution: 
For 2 years, the difference between compound interest (CI) and simple interest (SI) is given by:

Difference = P × (r/100)t
where, 
P = Principal amount
r = interest rate = 12%
Time = t = 2 years
Difference = 36 Taka

So, we get, 
36 = P × (12/100)2
36 = P × (144/10000)
∴ P = 2500 Taka

Question: Three workers can do a job in 12 days. Two of the workers work thrice as fast as the third. How long would it take one of the faster workers to do the job himself?

Solution:
Let,
The first and second worker need P days.
Third worker needs 3P days.

ATQ,
(1/P) + (1/P) + (1/3P) = 1/12
⇒ (3 + 3 + 1)/3P = 1/12
⇒ 7/3P = 1/12
⇒ 3P = 84
⇒ P = 84/3
∴ P = 28

So the one faster worker alone will take 28 days to complete the job.

Question: How many permutations of the letters of the word BALLOON are there?

Solution:
BALLOON শব্দটিতে মোট অক্ষর সংখ্যা = 7 টি
যার মধ্যে,
L আছে 2 বার
O আছে 2 বার

∴ Permutations = 7!/(2! × 2!)
= (7 × 6 × 5 × 4 × 3 × 2)/(2 × 2)
= 7 × 6 × 5 × 2 × 3 
= 1260

Speed downstream = (22 + 4) = 26 kmph
Time = 24 minutes
= 24/60 hour = 2/5 hour
Distance travelled = Time × speed
= (2/5) × 26 = 10.4 km

Question: ∠P and ∠Q are complementary to each other. If angle P = 20° + 4x and angle Q = 6x, find the value of ∠Q.

Solution:
Here, ∠P = 20° + 4x and ∠Q = 6x

For complementary angles,
∠P + ∠Q = 90°
⇒ (20° + 4x) + 6x = 90°
⇒ 20° + 10x = 90°
⇒ 10x = 90° - 20°
⇒ 10x = 70°
⇒ x = 70°/10 = 7°

∴ ∠Q = 6x = 6 × 7° = 42°

সকাল আটটা থেকে দুপুর দুইটা পর্যন্ত ঘড়ির ঘণ্টার কাটা ছয় ঘর সামনে যায়, অর্থাৎ = 360/12 × 6 ° = 180° ঘুরে