ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Let the average weight of the 59 students be A.
So the total weight of 59 of them will be = 59 × A = 59A

The questions state that when the weight of this student who left is added,
the total weight of the class=59A + 45
When this student is also included, the average weight decreases by 0.2 kgs

(59A + 45)/60 = (A - 0.2)
⇒ 59A + 45 = 60A - 12
⇒ 45 + 12 = 60A - 59A
⇒ A = 57.

The average of 6 players = X
Average increases by 10, when the seventh player makes a score of 112 runs.
Therefore, average of 7 players = X + 10
Average =Sum of Scores/Number of Players -------(1)
Here, average = X, number of players = 6
Hence,
Sum of scores = 6X
Score of 7 players = (Score of 6 players + score of 7 player) = (6X + 112)------- (2)
Total average = (X + 10) --------- (3)
Substitute (2) and (3), in (1)
(X + 10) = (6X + 112)/7
⇒ 7X + 70 = 6X + 112
⇒ 7X - 6X = 112 - 70
⇒ X = 42.

Average = Sum of quantities/Number of quantities
1) Sum of observations = Average x No. of observations
= 46 x 40 = 1840

2) Correct sum = Sum of observations + (38 – 33)
= 1840 + (5)
= 1845.
Corrected Mean Value = Corrected Sum/No. of Observations
= 1845/40
= 46.125

(1) First find total age of 5 members 2 years ago
(2) Present age of 5 members
(3) Total age of 6 members
(4) Age of baby = Total age of 6 members - Present age of 5 members

We know that,
Average = Sum of Quantities/Number of Quantities

1) First calculate total age of 5 members 2 years ago = (Average age of 5 members x number of members)
First calculate total age of 5 members 2 years ago = (16 x 5) = 80 years

2) Calculate the present age of 5 members
2 years ago, their total age was 80 years. Present age can be calculated as follows:
Present age of 5 members = [80 + (2 x 5)] = 90 years

3) Calculate total age of 6 members considering baby = (16 x 6 ) = 96 years

4) Age of baby = (96 – 90)
= 6 years.

Average of 11 numbers = 30

Step 1: Calculate total of 11 numbers by multiplying it by average value 30 = 11 x 30 = 330
Step 2: Calculate the total of the first six members by multiplying it by average value 17.5 = 17.5 x 6 = 105
Step 3: Calculate the total of the last six members by multiplying it by average value 42.5 = 42.5 x 6 = 255

Therefore,
we can find the sixth number by adding the value of the first six and last six numbers and subtracting it from the total value of 11 numbers.
Sixth number = (105 + 255) - 330
= 30.

Average = Sum of Quantities/Number of Quantities

1) First calculate the total age of 40 students
Total age of 29 students = ( Average age x No. of students)
= (20 x 29) = 580 years

2) Average age of 29 students + 1 teacher = 20 years + 3 months = 81/4 years

3) Finally the total age of 29 students + 1 teacher = 81/4 × 30 = 607.5 years

Therefore, age of teacher = (Total age of 30 members - Total age of 29 students)
= (607.5 – 580)
= 27.5 years.

Given: Average weight of batch A = 40 kg,
average weight of batch B = 35 kg

1) First find the total weight of all students
- Weight of batch A = (36 x 40) = 1440
- Weight of batch B = (44 x 35) = 1540

Total weight of all students = (1440 + 1540) = 2980 kg

2) Find average weight of whole class
(Batch A + Batch B) students = (36 + 44) = 80 students
Average = Total weight of all the students/No. of Students
= 2980/80
= 37.25 kg.

Consider the consecutive even numbers as : x, (x + 2), (x + 4) and (x+ 6)
Average = Sum of Quantities/Number of Quantities
{x + (x + 2) + (x + 4) + (x + 6)}/4 = 27
⇒ (4x + 12)/4 = 27
⇒ x + 3 = 27
⇒ x = 27 - 3
⇒ x = 24.

Therefore,
Largest number = (x + 6) = (24 + 6) = 30
Smallest number = 24.
Hence, the answer is 30.

Average of 15 numbers = 15, Average of 5 numbers = 14, Average of 9 numbers = 16

Average = Total numbers/Number of Numbers
15 = Total numbers/15
Therefore, total numbers = 15 x 15 = 225.

Middle number = (Total numbers) – [(Average of 5 num x no of num) + ( Average of 9 num x no of num)]
= (225) – [(14 x 5) + (16 x 9)]
= (225) – (214)
= 11.

We know,
Average = Sum of Quantities/Number of Quantities

Here,
Number of quantities = Number of students in each batch

As average marks of students are given, calculate total marks of each batch first. So total marks for
Batch 1 = (40 x 45) = 1800
Batch 2 = (50 x 55) = 2750
Batch 3 = (60 x 70) = 4200

Sum of marks = (1800 + 2750 + 4200) = 8750

Therefore,
Required Average = (Some of works)/(Total number of students in each batch)
= 8750/(40 + 50 + 60)
= 8750/150
= 58.33

Let daughter’s age be x and father’s age be 3x.
Father's age is 3 times more aged than his daughter,
therefore father’s present age = x + 3x = 4x
After 5 years,
The father’s age is 3 times more than his daughter's age.
(4x + 5) = 3 (x + 5)
(4x+5) = 3 (x+5)
(4x + 5) = 3 (x + 5)
x = 10

After 5 years it was (4x + 5),
then after further 5 years, father’s age = (4x +10) and daughter’s age = (x + 10)
(4x + 10)/(x + 10)
= [{(4 × 10) + 10}/(10 + 10)]
= 50/20
= 5/2
= 2.5
After further 5 years, the father will be 2.5 times of daughter’s age.

If ages in the numerical are mentioned in the ratio A : B, then A: B will be Ax and Bx
1) At present : Ratio of their ages = 5 : 3. Therefore, 5 : 3 will be 5x and 3x.
Rohan's age 4 years ago = 5x – 4
Rahul's age after 4 years = 3x + 4

2)Ratio of Rohan’s age 4 years ago and Rahul's age after 4 years is 1: 1
Therefore,
(5x - 4)/(3x + 4) = 1/1
⇒ 5x - 4 = 3x + 4
⇒ 5x - 3x = 4 + 4
⇒ 2x = 8
⇒ x = 4
Solving, we get x = 4

3)We are asked to find the ratio between Rohan's age 4 years hence and Rahul's age 4 years ago.
Rohan's age : (5x + 4)
Rahul's age: (3x – 4)
Ratio of Rohan's age and Rahul’s age
(5x + 4)/(3x - 4) = 24/8
(5x + 4)/(3x - 4) = 3/1
(5x + 4) : (3x - 4) = 3 : 1.

We are given that,
age of mother 10 years ago was 3 times the age of her son
So,
let the age of the son be x and as the mother's age is 3 times the age of her son, let it be 3x, three years ago.
At present,
Mother's age will be (3x + 10) and son's age will be (x + 10)
After 10 years,
Mother's age will be (3x + 10) +10 and son’s age will be (x + 10) + 10

Mother's age is twice that of son
(3x + 10) +10 = 2 [(x + 10) + 10]
(3x + 20) = 2[x + 20]
Solving the equation, we get x = 20
We are asked to find the present ratio.
(3x + 10) : (x + 10) = 70 : 30
(3x + 10) : (x + 10) = 7 : 3.

Let the age of Robin be y
Roman is 15 years older than Robin = (y + 15).
So Roman's age 5 years ago = (y + 15 – 5)
Robin's age before 5 years = (y – 5)

5 years ago, Roman is 3 times as old as Robin
(y + 15 – 5) = 3 (y – 5)
(y + 10) = (3y – 15)
2y = 25
y = 12.5
Robin's age = 12.5 years
Roman's age = (y + 15)
= (12.5 + 15)
= 27.5 years.

Let,
Jashim's present age be x
Jashim's age before 5 years = (x - 5)
Jashim's age after 10 years = (x + 10)
We are given that,
Jashim's age after 10 years (x + 10) is 5 times his age 5 years back (x – 5)

Therefore,
(x + 10) = 5 (x – 5)
⇒ x + 10 = 5x – 25
⇒ 4x = 35
⇒ x = 8.75 years.

According to the question,
A + B = 60,
A = 2B
2B + B = 60
⇒ B = 20 then A = 40.

5 years,
their ages will be 45 and 25.
Sum of their ages = 45 + 25
= 70.

Let present ages of A, B, and C be 4x, 7x, and 9x respectively.
(4x − 8) + (7x − 8) + (9x 8) = 56
20x = 80
x = 4

Hence present ages of A, B, and C are (4 × 4), (7 × 4), and (9 × 4)
That is 16, 28, and 36.

The present age of the son is P, he was born P years ago.
The age of the man was: (48 - P).

His age when the son was born should be equal to 2/3 of P.
(48 - P) = (2/3)P
5P = 144
⇒ P = 28.8 years.

Let the present age of the person be x years.

According to the question,
Then, 3(x + 3) - 3(x - 3) = x
⇒ 3x + 9 - 3x + 9 = x
⇒ 18 = x
Hence the present age of the person was 18 years.

Sum of ages of A and B, 3 Years ago =(18 x 2) = 36 Years
Sum of ages of A, B, and C,
Now = (22 x 3) = 66 Years
Sum of ages of A and B,
Now = (36 +6)Years = 42 Years

Therefore,
C's age = (66 - 42)Years = 24 Years.

Let C's age be x years.
Then, B's age = 2x years.
A's age = (2x + 2) years.

According to the question,
(2x + 2) + 2x + x = 27
5x = 25
⇒ x = 5
Hence, B's age = 2x = 10 years.

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

The present age of Tomal = 5
Present age of Razib = 5 - 2 = 3
Let present age of Akib = x
Then, present age of Razib = (x - 6)/18

According to the question,
(x - 6)/18 = 3
x - 6 = 54
x = 54 + 6
x = 60 years.

Sum of 18 observations = 18 x 124
=2232

Correct sum of 18 observations = 2232 - 46 - 82 + 64 + 28
= 2196
Therefore,
Correct average =2196/18
= 122.

Total age of grandparents = 67 × 2
Total age of parents = 35 × 2
Total age of grandchildren = 6 × 3

Total age of all persons = 67 × 2 + 35 × 2 + 6 × 3 = 222
Total number of persons = 2 + 2 + 3 = 7

The average age of the family
= 222/7
= 31(5/7)

Suppose the average expenditure was Tk. a.
Then total expenditure = 35a
When 7 more students join the mess, total expenditure = 35a + 42
Now, the average expenditure= (35a + 42)/(35 + 7)

Now, we have,
(35a + 42)/42 = (a - 1)
⇒ 35a + 42 = 42a – 42
⇒ 7a = 84
⇒ a = 12
Thus the original expenditure of the mess = 35 x 12 = 420.

Total salary of 65 workers = 65 × 5680 = Tk. 369200
Total salary of 31 workers = 31 × 2356 = Tk. 73036
Total salary of 23 workers = 23 × 4589 = Tk. 105547
No. of remaining workers = 65 - 31 – 23 = 65 – 54 = 11
Total Salary of 11 workers = 369200 – 73036 – 105547
= 369200 – 178583
= Tk. 190617
Required average = 190617/11
= Tk. 17328.81
Hence, the required average is Tk. 17328.81

Let the number of female students be x
Let the weight of female students = 57x
Number of male students = 66
Total weights of male students = 65.9 × 66
The average weight of all the students = 60.3 kg
Total weights of all the students = 60.3 (66 + x)

According to the given information,
Then,
⇒ 60.3 (66 + x) = 66 × 65.9 + 57x
⇒ 60.3 × 66 + 60.3x = 66 × 65.9 + 57x
⇒ 60.3x - 57x = 66 (65.9 - 60.3)
⇒ 3.3x = 66 × 5.6

∴ x = (66 × 5.6)/3.3
⇒ x = 2 × 56
⇒ x = 112