Bank Math

Let the base of the triangle is x and height of the triangle is y
Area of right triangle = 1/2 × xy = 6
∴ xy = 12

According to the Pythagorean Law,
x² + y² = 5²
(x + y)² - 2xy = 25
(x + y)² = 25 + 24 = 49
(x + y) = 7

So, the perimeter = 7 + 5 = 12 meters

Question: In how many ways can 8 football players be divided into two teams with an equal number of players?

Solution:
8 জন খেলোয়ার কে সমান দুটি ভাগে ভাগ করলে 8/2 = 4 জন করে দল গঠন করা যাবে।
 
তাহলে, 8 জন থেকে 4 জন করে নিয়ে পাই,  ⁸C₄ = 8!/[4!(8 - 4)!] 
= (8 × 7 × 6 × 5 × 4!)/(4! × 4!)
= (8 × 7 × 6 × 5)/(4 × 3 × 2)
= 70 

∴ দল গঠন করার উপায় = 70/2 = 35

Question: If a - b = 2 and a² + b² = 20, then the value of a + b will be?

Solution:
We know, (a - b)² = a² - 2ab + b²
⇒ (a - b)² = a² + b² - 2ab ----------------(1)

Given,
a - b = 2
⇒ (a - b)² = 4
Also given, a² + b² = 20

Substitute the values in equation (1): 
20 - 2ab = 4
⇒ 2ab = 16
⇒ ab = 8

Now, (a + b)² = a² + b² + 2ab
= 20 + 16
= 36
∴ a + b = ±6

Question: The ratio of the number of boys and girls in a college is 7 ∶ 8. If the percentage increase in the number of boys and girls be 20% and 10% respectively, what will be the new ratio?

Solution: 
ধরি,
কলেজে ছেলে এবং মেয়ের সংখ্যা যথাক্রমে 7x এবং 8x

∴ ছেলেদের বর্ধিত সংখ্যা = 120% × 7x 
= (120/100) × 7x 
= (6/5) × 7x 
= (42x)/5

এবং, মেয়েদের বর্ধিত সংখ্যা = 110% × 8x
= (110/100) × 8x
= (11/5) × 4x
= (44x)/5

 ∴ বর্ধিত সংখ্যার অনুপাত (New ratio) = (42x)/5 ∶ (44x)/5
= 21 ∶ 22

Simple interest I = pnr/100
A/Q, (400 × 100 × n)/100 = (1000 × 4 × 4)/100
Or, 4000n = 16000
Or, n = 4

Question: How many 'P's are there in the following sequence which are immediately preceded by 'R' and immediately followed by 'Q'?
R P Q L P R Q R P L R P Q R L P Q R P Q L R L P R P Q L R P

Solution:
প্রদত্ত অনুক্রমটি বিশ্লেষণ করি:
R P Q L P R Q R P L R P Q R L P Q R P Q L R L P R P Q L R P

- অনুক্রমটি বিশ্লেষণ করলে দেখা যায়, 'RPQ' বিন্যাসটি মোট 4 বার রয়েছে।

Let the ten's digit be x.
Then, the unit's digit = 2x + 1.
[10x + (2x + 1)] - [{10 (2x + 1) + x} - {10x + (2x + 1)}] = 1
⇒ (12x + 1) - (9x + 9) = 1
⇒ 3x = 9, x = 3.
So, ten's digit = 3 and unit's digit = 7.
Hence, original number = 37.

Question: One pipe can fill a tank four times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the faster pipe alone will be able to fill the tank in:

Solution:
Let,
the slower pipe alone fill the tank in x minutes.
Then, Faster pipe alone will fill it in x/4 minutes.

ATQ,
(1/x) + (4/x) = 1/36
⇒ 5/x = 1/36 
∴ x = 180

The slower pipe alone fill the tank in 180 minutes.
the faster pipe alon will be able to fill the tank in = (180 ÷ 4) minutes.
= 45 minutes

First, identify the number that is multiple of 3 more than 100.
That type of number is 102.
So, 102//3 = 34.
Second, we have to identify the number that is multiple of 3 but nearest less than 198.
Now, 198/3 = 66.
Hence, the answer is (66 - 34) + 1 = 33

Question: If p and q are positive integers and (p + q) is an even number, then (p² + q²) will be always divisible by-

Solution:
In this problem, put any even positive value for both p and q,

For example m = 6 and n = 4
∴ (p² + q²) = (6² + 4²)
= 36 + 16
= 52 is always divisible by 2.

Question: The wheel of a scooter has a diameter of 140 cm. How many revolutions per minute must the wheel make to maintain a speed of 132 km/h?

Solution:
Distance travelled by wheel in one revolution = circumference of wheel
= (22/7) × 140 = 440 cm.

And
Speed of scooter = 132 km/hr = (132 × 1000 × 100)/60 cm/min = 220000 cm/min.

∴ Revolutions per minute = Distance covered per minute/Distance per revolution
= 220000/440 = 500

So the answer is indeed 500 revolutions per minute.

Let speed upstream be x km/hr
Then, speed downstream = 3x km/hr.

Speed in still water = 1/2 (3x + x) kmph = 2x km/hr.

∴ 2x = 28/3
x = 14/3 km/hr;
Speed downstream = 14 km/hr

Hence, speed of the current
1/2 {14 - (14/3)} km/hr
= 14/3 km/hr
= 4(2/3) km/hr.

Question: Water : Dam :: Trade: ?

Solution:
উত্তর: ঘ) Trade Policy
এখানে প্রথম জোড়ার সম্পর্কটি হলো নিয়ন্ত্রণ ব্যবস্থা।
- একটি ড্যাম (বাঁধ) পানির প্রবাহকে নিয়ন্ত্রণ করে, নিয়ন্ত্রিতভাবে ছাড়ে, বন্যা রোধ করে, পানি সঞ্চয় করে এবং প্রয়োজনমতো ব্যবহারের জন্য নিয়ন্ত্রণ করে। 
- অর্থাৎ, Dam হলো Water-এর নিয়ন্ত্রক/নিয়ন্ত্রণকারী ব্যবস্থা।

একইভাবে, Trade : Trade Policy এর ক্ষেত্রে:
- Trade Policy (বাণিজ্য নীতি) হলো সরকারের তৈরি করা নিয়ম-কানুন, যেমন—শুল্ক (tariff), কোটা (quota), আমদানি-রপ্তানি নিষেধাজ্ঞা, বাণিজ্য চুক্তি ইত্যাদি। 
- এই নীতিগুলো বাণিজ্যের (Trade) প্রবাহকে নিয়ন্ত্রণ করে, দেশীয় বাজার রক্ষা করে, আমদানি-রপ্তানির পরিমাণ নিয়ন্ত্রণ করে এবং বাণিজ্যকে পরিচালিত করে।
- অর্থাৎ, Trade Policy হলো Trade-এর নিয়ন্ত্রক/নিয়ন্ত্রণকারী ব্যবস্থা।

Question: Rahim invested 20,000 Tk. in two 4 years term. He received 6% for the first term and 9% for the second term. What is the average rate of interest did he receive?

Solution:
In first term, he received = 20,000 × 4 × 6%
= 4,800 Tk.
In second term, he received = 20,000 × 4 × 9%
= 7,200 Tk.

So, total interest after 8 years, I = 4,800 + 7,200 = 12,000
P = 20,000
n = 8
r = ?

We know,
I = Pnr/100
∴ r = (100 × I)/Pn
= (100 × 12,000)/(20,000 × 8)
= 7.5%

∴ The average rate of interest Rahim received = 7.5%

Let the length of the train be x
∴ x/20 = x + 250/45
⇒ 5x = 1000
⇒ x = 200 m

Question: A ladder 25 m long leans against a wall. The foot of the ladder is 7 m from the wall. How high up the wall does the ladder reach?

Solution:
Given that,
The ladder is the hypotenuse = 25 m
The distance from the foot of the ladder to the wall is one leg = 7 m

Applying the Pythagorean theorem,
Hypotenuse² = a² + h²
⇒ 25² = 7² + h²
⇒ h² = 625 - 49
⇒ h² = 576
⇒ h = √576
∴ h = 24

So the ladder reaches 24 m high up the wall.

Question: Compound interest on a certain sum for 2 years at 10% per annum is Tk. 630. What would be the simple interest at the same rate and for the same time?

Solution:
Let, Principle = P
Compound Principle, C = P(1 + r)n
= P(1 + 10/100)2

ATQ,
C - P = 630
⇒ P(1 + 10/100)² - P = 630
⇒ P{(11/10)² - 1} = 630
⇒ P{(121/100) - 1} = 630
⇒ P(121- 100)/100 = 630
⇒ P(21/100) = 630
⇒ P = (630 × 100)/21
∴ P = 3000

Again,
I = Pnr
= 3000 × 2 × (10/100)
= 600

∴ The simple interest would be Tk. 600. 

Question: A bag contains 5 red, 3 green, and 4 blue balls. Two balls are drawn without replacement. What is the probability that the first ball is green and the second ball is red or blue?

Solution:
Given that,
Red balls = 5
Green balls = 3
Blue balls = 4
∴ Total balls = 5 + 3 + 4 = 12

We know Probability = Favorable outcomes/Total outcomes

Now, First ball is Green = 3/12 = 1/4

If first is green, then the remaining balls = 11
Second ball is Red or Blue = (5 + 4)/11 = 9/11

∴ Required probability = (1/4) × (9/11)
= 9/44

Let the time taken to fill the cistern by 3 taps A, B and C be X, Y, and Z minutes respectively.
Then the short cut formula for,
Time taken to fill the tank when all the pipes are opened = XYZ/(XY + YZ + ZX) minutes
Here, X = 9 minutes, Y = 6 minutes and Z = 3 minutes.
Now the required time = {(9) × (6) × (3)}/{(9 × 6) + (6 × 3) + (3 × 9)} minutes
= (9 × 6 × 3)/(54 + 18 + 27)
= (9 × 6 × 3)/9{6 + 2 + 3}
= 6 × 3/(6 + 2 + 3)
= 18/11
= 1(7/11) minutes
Hence the answer is 1(7/11) minutes.

Question: What is the probability of having 53 Sundays in a leap year?

Solution:
Leap year = 366 Days
= {(52 × 7) + 2} Days
= 52 Weeks + 2 Days

∴ Probability of having 53 Sundays = 2/7

P(first letter is not vowel) = 2/4
P(second letter is not vowel) = 1/3
So, probability that none of letters would be vowels is = 2/4 × 1/3
= 1/6

Question: The average of x and y is 45, and the average of y and z is 50. If y = 42, then find the value of (x + z).

Solution:
Given y = 42

Average of x and y = 45
∴ x + y = 45 × 2
⇒ x + y = 90
⇒ x + 42 = 90
⇒ x = 90 - 42
⇒ x = 48

Average of y and z = 50
∴ y + z = 50 × 2
⇒ y + z = 100
⇒ 42 + z = 100
⇒ z = 100 - 42
⇒ z = 58

Now, x + z = 48 + 58 = 106

Let the sheet be folded along its breadth and its perimeter = 48cm

Therefore, (l/2 + b) = 48 ......(i)
Now, let the sheet be folded along its length, and the perimeter = 66cm
(l + b/2)= 66 ........(ii)
Solving (i) and (ii), we get,
l = 56 cm, b = 20 cm
Area = l × b = 56 cm × 20 cm
⇒ Area = 1120 cm²

3^x + 3^x + 3^x
= 3.3^x
= 3¹.3^x
= 3^{x + 1}

Question: If the average of x and y is 70 and the average of y and z is 90. What is the value of z - x?

Solution: 
(x + y)/2 = 70
∴ x + y = 140 .........(1)

Again,
(y + z)/2 = 90
∴ y + z = 180 .........(2)

From (2) - (1) we get,
y + z - x - y = 180 - 140
∴ z - x = 40

Question: The marked price of a shirt was Tk. 2400. A customer bought it for Tk. 1620 after getting two successive discounts. The first discount was 25%. What was the second discount rate?

Solution:
Price after first discount (25%),
= 2400 - (25% of 2400)
= 2400 - (2400 × 0.25)
= 2400 - 600
= Tk. 1800

Amount of second discount = 1800 - 1620 = Tk. 180

∴ Second discount rate = (Second discount amount/Price after first discount) × 100
= (180/1800) × 100
= 0.10 × 100
= 10%

∴ The second discount rate is 10%.

x = 80%
⇒ x = 80/100 = 4/5
⇒ x² = 16/25 = 16/25 × 100% = 64%
x is larger than x² (in percentage) = (80% - 64%) = 16%
x is larger than x² = (80% - 64%) = 16%
x is larger than x² (in percentage) = (16/64 ×100) = 25%

Question: If the nth term of an arithmetic progression is 5n + 3, then what is the common difference?

Solution:
The nth term of an arithmetic progression is T_n = 5n + 3

n = 1 then, T₁ = 5 × 1 + 3 = 8
n = 2 then, T₂ = 5 × 2 + 3 = 13
n = 3 then, T₃ = 5 × 3 + 3 = 18
n = 4 then, T₄ = 5 × 4 + 3 = 23

Common difference, T₂ - T₁ = 13 - 8 = 5
T₄ - T₃ = 23 - 18 = 5

∴ The common difference is 5.

16 years ago,
Let, Tania (T) = x years and Grandfather (G) = 8x years.
After 8 years from now,
T = (x + 16 + 8) years and G = (8x + 16 + 8) years.

8x + 24 = 3(x + 24)
⇒ 5x = 48
8 years ago,
T/G = (x + 8)/(8x + 8)
= {(48/5) + 8}/{8 × (48/5) + 8}
= (88/5)/(424/5)
= 88/424
= 11/53

Let, the number y
According to the question, 
y² + 12y = - 27
or, y² + 12y + 27 = 0
or, y² + 3y + 9y + 27 = 0
or, y(y + 3) + 9(y + 3) = 0
or, (y + 3)(y + 9) = 0
or, y = - 3, - 9
The smaller possible value of this number = - 9

Note that bigger number with negative value is actually the smaller number.

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

Solution:
Here,
∠M = 20° + 4x and ∠N = 6x

For complementary angles,
∠M + ∠N = 90°
⇒ (20° + 4x) + 6x = 90°
⇒ 20° + 4x + 6x = 90°
⇒ 20° + 10x = 90°
⇒ 10x = 90° - 20°
⇒ 10x = 70°
∴ x = 7°

So, ∠N = 6 × 7° = 42°

Question: A sum of 10,000 Taka is invested at 8% per annum. If the interest is compounded quarterly, the amount after 1 year will be:

Solution: 
The annual rate is 8%.
Since interest is compounded quarterly, the quarterly rate is:
8%/4 = 2% per quarter

Time period is 1 year, meaning there are 4 quarters.

We apply 2% interest every quarter.

First Quarter:
New Amount=10,000 + (2% of 10,000) = 10,000+200= 10,200

Second Quarter:
New Amount=10,200 + (2% of 10,200) =10,200+204= 10,404

Third Quarter:
New Amount=10,404 + (2% of 10,404) =10,404+208.08= 10,612.08

Fourth Quarter:
New Amount=10,612.08 + (2% of 10,612.08) =10,612.08+212.24= 10,824.32

So, The final amount after 1 year is 10,824 Taka.

Question: The length of two chords AB and AC of a circle are 8 cm and 6 cm and ∠BAC = 90º, then the radius of circle is-

Solution:
Given that,
Chord AB = 8 cm
Chord AC = 6 cm
∠BAC = 90°
Since ∠BAC = 90°, triangle ABC is right-angled at A. The hypotenuse BC is the diameter of the circle.

We know, 
BC = √(AB² + AC²)
⇒ BC = √(8² + 6²) = √(64 + 36) = √100
∴ BC = 10 cm

Radius = half of diameter
∴ r = BC/2 = 10/2 = 5 cm

Question: A certain sum of money consists of 30 coins, some of which are 10 paisa coins and the rest are 5 paisa coins. If the total value of the coins is Tk. 2, what is the number of 10 paisa coins?

Solution:
Let the number of 10 paisa coins = x
Then the number of 5 paisa coins = 30 - x

According to the question,
10x + 5(30 - x) = 200 (since Tk. 2 = 200 paisa)
⇒ 10x + 150 - 5x = 200
⇒ 5x = 50
⇒ x = 10

∴ Number of 10 paisa coins = 10

Question: A fair coin is tossed 4 times. What is the probability of getting exactly 3 heads?

Solution:
For 4 tosses, the total number of possible outcomes is = 2⁴ = 16
The number of ways to choose 3 Heads out of 4 tosses is = ⁴C₃ = 4

∴ Probability of getting exactly 3 heads = 4/16 = 1/4

Let AB be the pole and BC be its shadow.
Consider θ is the angle of elevation of the Sun.



In right triangle ABC,

tan θ = AB/BC = h/√3h = 1/√3
tan θ = tan 30°
θ = 30°

Let x=log2⁷
=> 2x=7
which is only possible for irrational number

Let original average be x

then (12x+ 96) / 13 = x + 5
⇒ 12x + 96 = 13x + 65
⇒ x = 96 - 65 = 31
∴ His average after 13 innings = 31 + 5 = 36

HCF of the two numbers = 23
Since HCF will always be a factor of LCM, 23 is a factor of the LCM.
Given that other two factors in the LCM are 13 and 14.
Hence factors of the LCM are 23, 13, 14
So, numbers can be taken as (23 × 13) and (23 × 14)
= 299 and 322
Hence, largest number = 322.

The trapezium is a quadrilateral with one pair of parallel sides.
Let the length of the smaller side = 12x
Length of the bigger parallel side = 18x

Distance between parallel sides = 32 meters

Area of a trapezium = 1/2 × (sum of parallel sides) × distance between them.

So, as per the question
1500 = 1/2 × (12x +18x) × 30
⇒ 1500 = 1/2 × 30x × 30
⇒ 1500 = 30x × 15
⇒ 30x = 1500 /15
⇒ 30x = 100
⇒ x = 100 /30
= 3.3 meters.

Question: A, B and C together invest Tk. 20000 in a business. A invests Tk. 2000 more than B and B invests Tk. 3000 more than C. Out of a total profit of Tk. 6000, find the share of A.

Solution:
Let C's investment = x
Then, B invests Tk. 3000 more than C
∴ B = x + 3000

A invests Tk. 2000 more than B
∴ A = (x + 3000) + 2000 = x + 5000

Total investment,
A + B + C = 20000
⇒ (x + 5000) + (x + 3000) + x = 20000
⇒ 3x + 8000 = 20000
⇒ 3x = 20000 - 8000
⇒ 3x = 12000
∴ x = 4000

Now,
C = x = 4000
B = 4000 + 3000 = 7000
A = 4000 + 5000 = 9000

∴ Profit sharing ratio is,
A : B : C = 9000 : 7000 : 4000 = 9 : 7 : 4

Sum of the ratios = 9 + 7 + 4 = 20

∴ Share of A = 6000 × (9/20)
= 2700

So, the share of A is Tk. 2700.