ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Since the month begins with a Sunday, so there will be five Sunday in the month.
∴ Required average = {(510 × 5) + (240 × 25)}/30 = 8550/30 = 285
Answer: 285
Source: Quantitative Aptitude

Let A's age be 5x years.
Then, B's age = 3x years.
So,
(5x - 4)/3x + 4 = 1/1
⇒ 5x - 4 = 3x + 4
2x = 8
⇒ x = 4
∴ A's age 4 years hence/B's age 4 years ago = 5x + 4/3x - 4
= (5 × 4 + 4)/(3 × 4 - 4)
= 24/8
= 3/1
= 3:1
Answer: 3:1

Let there be x pupils in the class.
The total increase in marks = (x × 1/2) = x/2
∴ x/2 = (83 - 63)
⇒ x/2 = 20
⇒ x = 40
Answer: 40

Bubli's age = 14 years
⇒ Sakib's age = (14 - 9) years
= 5 years
Let the present age of Zayed Khan be x years.
∴ (x - 10)/14 = 5
⇒ x - 10 = 70
⇒ x = 80 years.
Answer: Zayed Khan age is 80 years.

Let the ten's digit be x and unit's digit be y.
Then, (10x + y) - (10y + x) = 63
⇔ 9 (x - y) = 63
x - y = 7.
There are several numbers like this, e.g. 70-07, 81-18 and 92-29. Thus, the correct answer is - ঘ) Can not be determined

Let the number be x.
Then, x/5 + 4 = x/4 - 10
⇔ x/4 - x/5 = 14
⇔ x/20 = 14
⇔ x = 20 × 14 = 280.
Answer: 280.

The average of 11 results = 60
∴ The total of 11 results = 60 × 11 = 660
Average of first six results = 58
∴ Total of first six results = 58 × 6 = 348
Average of last six results = 63
∴ Total of last six results = 63 × 6 = 378
∴ sixth results = total of first and last six results - total of 11 results 
                     = (348 + 378) - 660
                     = 726 - 660
                     = 66
Answer: 66

Let, the numbers be a and b where a > b.
According to the question,
a - b = 10 ....... (i)
And (a + b)/5 = 8
By cross multiplying, we get
⇒ a + b = 40 ...... (ii)
By subtracting equation (ii) from (i) we get
2b = 40 - 10 = 30
⇒ b = 30/2 = 15

And from (i)
a = 10 + 15 = 25
Answer: 15

Let,
Fahim's age = x
Bilash's age = y and
Wasim's age = z.

According to the given information
Age of Bilash - Age of Fahim = Age of Fahim - Age of Wasim
⇒ y - x = x - z
⇒ 2x = y + z ......(i)
Also, y + z = 66
from (i) x = 33 years

but, y + z = 66 means, there can be many sets of values that will satisfy the equation.
01 - 65
02 - 64
... ... ...
32 - 34

We want to determine x - z but as we can't get a fixed value for z; hence, the answer can not be determined.

16 years ago, let V = x years and G = 8x years.
After 8 years from now, V = (x + 16 + 8)years and G = (8x + 16 + 8) years.
∴ 8x + 24 = 3 (x + 24)
⇒ 8x - 3x = 72 - 24
⇒ 5x = 48
8 years ago, V/g = (x + 8)/(8x + 8)
= {(48/5) + 8}/{8 × (48/5) + 8}
= (48 + 40)/(384 + 40) = 88/424
= 11/53
Answer: none of these.

Let the numbers be x and (x + 16)
Then, x/3 - (x + 16)/7 = 4
⇔ 7x - 3(x +16) = 84
⇔ 4x = 84 + 48 = 132
⇔ x = 132/4 = 33
hence, the numbers are 33 and 49
Answer: 33 and 49

Let A, B, C represent their respective weights. Then, we have:
A + B + C = (45 x 3) = 135 .... (i)
A + B = (40 x 2) = 80 .... (ii)
B + C = (43 x 2) = 86 ....(iii)
Adding (ii) and (iii), we get: A + 2B + C = 166 .... (iv)
Subtracting (i) from (iv), we get : B = 31
∴ B's weight = 31 kg.

The average age of five members is 27
Total age = 27× 5 = 135
After excluding one person, the new average = 27 - 2 = 25
New total age = 25× 4 = 100
Then the age of excluded person = Total age - New total age
= 135 - 100
= 35
Hence the required answer is 35.

Let x be the number of students in the class and y be the average weight of the class
Now according to question,
(xy + 50)/(x + 1) = y + 1
x + y = 49 .......... (i) Again,
(xy + 50 + 50)/(x + 2) = y + 1.5
1.5x + 2y = 97.......... (ii) From equation (i) and (ii), we get y = 47
Answer: 47

Let,
The son's present age is x years.
Then, the man's present age is = (x + 24)
∴ (x + 24) + 2 = 2(x+ 2)
⇒ x + 26 = 2x + 4
⇒ x = 22

Sum of the present age of husband, wife and child
= (23 × 2 + 5 × 2) + 1 = 57 years
∴ Required average = (57/3) years
= 19 years

According to Rakib 17th, 18th or 19th ...... (i)
According to her sister 19th, 20th, 21st or 22nd ......(ii)
From (i) and (ii) ⇒ 19th
Answer: 19

Let,
The number of boys and girls in the class are 3x and x respectively.
Let the average score of the girls be y.
Then, 3x(A + 1) + xy = (3x +x)A
⇒ 3(A + 1) + y = 4A
⇒ 3A + 3 + y = 4A
⇒ y = A - 3

Age decreased = (5 × 3) years
= 15 years
So the required difference = 15 years

Mobin's age:Xam's age = 4:5 = 1:5/4
Mobin's age:Faisal's age = 5:6 = 1:6/5
Let, Mobin's age be x years. Then, Xami's age = 5x/4
And, Faisal's age = 6x/5 years
∴ x + 5x/4 + 6x/5 = 69
⇒ 20x + 25x + 24x = 69 × 20
⇒ 20x + 25x + 24x = 1380
⇒ 69x = 1380
⇒ x = 1380/69
= 20
Xami's age = 5x/4 = (5 × 20)/4
= 25 years

P + C + M = C + 120
⇒ P + M = 120
∴ Required average = (P + M)/2 = 120/2
= 60

Correct sum = 36 × 100 + 90 - 40
= 3650
Correct average = 3650/100 = 36.5
Error = (36.5 - 36) = 0.5
∴ Error% = {(0.5/36.5) × 100}% = (100/73)%
= 1.36%

Let the present age of A be a years and that of B be b years
Then, 4 years ago,
A's age = (a - 4)
B's age = (b - 4)
Now, according to the given information in question,
{(a - b)/2}/4(b - 4) = 5/12 or (a - 4)/2(4b - 16) = 5/12 or (a - 4)/(4b - 16) = 5/6
By cross multiplying we get
or, 6a - 24 = 20b - 80
or, 6a - 20b = -56
or, 10b - 3a = 28
After 8 years, 
(a + 8)2 + 2 = b + 8
or, a/2 + 4 + 2 = b + 8
or, b - a/2 = -2
or, 2b - a = -4 .......(i)
a = 2b + 4 ......(ii)
Putting the value of a in equation (i), we get
10b - 3(2b + 4) = 28
or, 4b = 40
∴ b = 10
Hence, the present age of B is 10 years.

A's age = 44 × (6/11) years = 24 years and
B's age = (44 - 24) years = 20 years.
Ratio of their ages after 8 years = (24 + 8)/(20 + 8)
= 32/28
= 8/7
= 8:7