Bank Math

Compound interest for the 4th year = 3600 + (3600 × 8 × 1)/100.
= 3600 + 288.
= Tk. 3888.

Compound interest for the 5th year = 3888 + (3888 × 8 × 1)/100.
= 3888 + 311.04.
= Tk. 4199.04.

Difference between the compound interest in the 4th and 5th year = 4199.04 - 3888.
= Tk. 311.04.

ধরি, মুদি ব্যবসায়ী x টি ডিম কিনেছিল।
প্রশ্নমতে, 4 (x - 12) - 3x = 96
বা, 4x - 48 - 3x = 96
বা,  x = 96 + 48
বা, x = 144

Question: 60% of a smaller number is 6 less than 50% of a larger number. The larger number is 60 greater than the smaller one. The sum of these two number is-

Solution:
Let the smaller number be x and larger number be y
According to the question,
60% of x + 6 = 50% of y
⇒ (3x/5) + 6 = y/2
⇒ 6x + 60 = 5y
⇒ 6x - 5y = - 60 ................ (1)
and
y - x = 60 ----------- (2)

By using (1) and (2), we get
x = 240
y = 300

∴ The sum of these two number =(300 + 240) = 540

When Rahim meets Shafiq for the third time,
they together would have covered a Distance of 5d, i.e 5 × 30m = 150 m.

The ratio of Speed of Rahim and Shafiq = 2 : 1,
so the total distance traveled by them will also be in the ratio 2 : 1
as the Time is taken is constant.

So the Distance traveled by Rahim will be (2/3) × 150= 100 m.

Question: A man on tour travels first 60 km at 20 km/hr and the next 60 km at 30 km/hr. The average speed for the first 120 km of the tour is :

সমাধান:
প্রথম অংশের জন্য সময় = দূরত্ব/গতিবেগ
 = 60 কিমি/20 কিমি/ঘন্টা
= 3 ঘন্টা

দ্বিতীয় অংশের জন্য সময় = দূরত্ব/গতিবেগ
= 60 কিমি/30 কিমি/ঘন্টা
= 2 ঘন্টা

মোট অতিক্রান্ত দূরত্ব = 60 কিমি + 60 কিমি = 120 কিমি
মোট সময় = 3 ঘন্টা + 2 ঘন্টা = 5 ঘন্টা

∴ গড় গতিবেগ = 120 কিমি/5 ঘন্টা
= 24 কিমি/ঘন্টা

Let the son's present age be x years.
Then, man's present age = (x + 24) years
=> (x + 24) + 2 = 2(x + 2)
=> x + 26 = 2x + 4
So, x = 22

(4M + 6W) × 8 = (3M + 7W) × 10
=> M / W = 11 : 1
Now , (4 × 11 + 6 × 1) × 8 = 10 × 1 × T
=> T = 40 days

Commission for sale upto 6000 = 6000 × 5% = 300
Commission for sales over 6000 = 4000 × 8% = 320
Total commission = 620.

Question: P can build a wall in 24 days and Q can do it in 20 days. With help of R, they completed the work in 6 days. Find in how many days R alone can do the work.

Solution:
P can do 1/24 part per day
Q can do 1/20 part per day
R can do per day = (1/6) - {(1/24) + (1/20)}
= (1/6) - (11/120)
= (20/120) - (11/120)
= 3/40
Time taken by R alone = 40/3 days

Question: 15 buckets of water fill a tank when the capacity of each bucket is 16 liters. How many buckets will be needed to fill the same tank, if the capacity of each bucket is 12 liters?

Solution: 
total capacity of the tank is = (15 × 16) = 240 liters.

total buckets of 12 liters = 240/12 = 20 buckets

Question: If p and q are the roots of the equation 2x² − 9x + 7 = 0, then what is the value of (1/p) + (1/q)?

 
Solution:
Given equation:
2x² − 9x + 7 = 0
⇒ 2x² − 7x − 2x + 7 = 0
⇒ x(2x − 7) − 1(2x − 7) = 0
⇒ (x − 1)(2x − 7) = 0

So the roots are:
x = 1 = p
x = 7/2 = q

Now,
1/p + 1/q
= 1/1 + 1/(7/2)
= 1 + 2/7
= 9/7

A tap can fill a tank in 4 hours.
Therefore the tap can fill half the tank in 2 hours.

Remaining = 1/2

After half the tank is filled, three more similar taps are opened.
Hence, the total number of taps becomes 4.

Part filled by one tap in 1 hour = 1/4
Part filled by four taps in 1 hour = 4 × (1/4) = 1
i.e., 4 taps can fill the remaining half in 30 minutes.

Total time taken
= 2 hour + 30 minute = 2 hour 30 minutes.

Question: What is the smallest number of soldiers that can be arranged in groups of 12, 15, 18, and 20, and also arranged to form a perfect square?

Solution:
LCM of 12, 15, 18, and 20 is-
12 = 2 × 2 × 3
15 = 3 × 5
18 = 2 × 3 × 3
20 = 2 × 2 × 5

∴ LCM = 2 × 2 × 3 × 5 × 3

Since the soldiers are in the form of a solid square.
Hence, LCM must be a perfect square. To make the LCM a perfect square, we have to multiply it by 5,

∴ The required number of soldiers = 2 × 2 × 3 × 3 × 5 × 5
= 900

Question: The combined ages of 5 children, with each child 4 years older than the next, equal 70 years. How old is the oldest?

Solution:
Let the ages of children be x, (x + 4), (x + 8), (x + 12) and (x + 16) years.
Then, 
x + (x + 4) + (x + 8) + (x + 12) + (x + 16) = 70
5x + 40 = 70
5x = 70 - 40
⇒ 5x = 30
⇒ x = 6
∴ Age of the eldest child = x + 16 = 6 + 16 = 22 years.

Question: If log_m 128 = 7, then find the value of m.

Solution:
log_m 128 = 7
⇒ m⁷ = 128  [log_b A = C, then b^C = A]
⇒ m⁷ = 2⁷
∴ m = 2

Question: What word can be formed by arranging the letters of 'AENIPNOMU'?

Solution:
Given letters: A, E, N, I, P, N, O, M, U
By rearranging the letters 'AENIPNOMU', the name of a disease can be formed:
AENIPNOMU ⇒ PNEUMONIA

• PNEUMONIA is a serious respiratory disease that causes inflammation of the lungs, typically caused by bacterial or viral infection.

∴ The correct answer is — A name of a disease.

Question: Of three numbers, the average of the first and second numbers is 12 more than the average of the second and third numbers. What is the difference between the first and third numbers?

Solution: 
Let, these numbers are x, y and z respectively.

ATQ,
{(x + y)/2} - {(y + z)/2} = 12
⇒ {(x + y) - (y + z)}/2 = 12
⇒ (x + y - y - z)/2 = 12
∴ x - z = 24

∴ The difference between the first and the third number is = 24.

Question: If a + b = √13 and a - b = √5, what is the value of 8ab(a² + b²)?

Solution:
দেওয়া আছে,
a + b = √13
a - b = √5

আমরা জানি,
2(a² + b²) = (a + b)² + (a - b)²
4ab = (a + b)² - (a - b)²

এখন,
8ab(a² + b²) = (4ab) × 2(a² + b²)
= [(a + b)² - (a - b)²][(a + b)² + (a - b)²]
= [(√13)² - (√5)²][(√13)² + (√5)²]
= (13 - 5)(13 + 5)
= (8)(18)
= 144

Question: A stock increases in value by 20%. By what percent must the stock decrease to reach back to its former value?

Solution:
এখানে,
20% বৃদ্ধিতে মূল্য = (100 + 20) = 120 টাকা

120 টাকায় মূল্য কমাতে হবে 20 টাকা
∴ 1 টাকায় মূল্য কমাতে হবে 20/120
∴ 100 টাকায় মূল্য কমাতে হবে (20 × 100)/120
= 16.66 টাকা

Question: If Nita is 15 ahead in rank of Mita who ranks 13th from the last, then how many students are there in the class if Nita ranks 5th in order of merit?

Solution:
Let the total number of students = n

Now,
Mita’s position from the last = 13th
∴ Mita’s position from the front = n - 13 + 1 = n - 12

Again,
Nita’s position from the front = 5th.

And, Nita is 15 ranks ahead of Mita.
∴ Mita’s position = Nita’s position + 15.

According to the problem,
n - 12 = 5 + 15
⇒ n - 12 = 20
⇒ n = 20 + 12
∴ n = 32 

Therefore, the total number of students in the class = 32.

Question: If a^1/x = b^1/y = c^1/z and abc = 1, then find the value of x + y + z.

Solution: 
Let, a^1/x = b^1/y = c^1/z = k
a = k^x, b = k^y and c = k^z
abc = k^x × k^y × k^z = k^{(x + y + z)}

Given, abc = 1
k^{(x + y + z)} = k⁰
x + y + z = 0

Let x be the larger number and y be the smaller number
Therefore x - y = 20%of x
Now put the value of the smaller number in the equation
x - 20 = (20/100) × x
⇒ x - 20 = (1/5) × x

Here 5 is in the denominator.
As we bring 5 on the left side of the equation it will be multiplied by x - 20
Now the equation will be 5 × (x - 20) = x
⇒ 5x - 100 = x

Bring 100 on the right side of = and x on the left side of =
So it will become 5x - x=100
⇒ 4x = 100
⇒ x = 100/4
⇒ x = 25
Therefore the larger number is 25.

Question: Solution set of the inequality, 4x - 7 ≤ 2x + 5 is-

Solution:
Given that,
4x - 7 ≤ 2x + 5
⇒ 4x - 2x ≤ 7 + 5
⇒ 2x ≤ 12
⇒ x ≤ 6
∴ x ≤ 6

∴ Solution set of the inequality is (- ∞, 6]

To earn Tk. 10, money invested = Tk. 100.
To earn Tk. 12, money invested = Tk (100/10 × 12)
= Tk. 120.
Market value of Tk.100 stock = Tk. 120.

Question: If measures of the angles in a triangle are in the ratio of 1 : 3 : 5, then the degrees in the largest angle:
(Officer Cash 2022 অনুযায়ী)

Solution:
Given that,
The angles of a triangle are in the ratio 1 : 3 : 5
Let,
x, 3x, 5x

We know that,
Sum of angles in a triangle = 180°

Now
x + 3x +5x = 180°
9x = 180°
x = 180°/9 = 20°
∴ x = 20°

∴ Largest angle = 5x = 5 × 20 = 100°