সিনিয়র অফিসার নিয়োগ প্রস্তুতি (আর্কাইভ)

Question: Three numbers are in the ratio 2 : 3 : 5, and the sum of their squares is 608. Find the smallest number.

Solution:
Let the numbers be 2x, 3x, 5x

ATQ,
(2x)² + (3x)² + (5x)² = 608
⇒ 4x² + 9x² + 25x² = 608
⇒ 38x² = 608  
⇒ x² = 608/38
⇒ x² = 16 = 4²
∴ x = 4

∴ Smallest number = 2x = 2 × 4 = 8

Question: The average income of A, B and C is Tk. 12000 per month and average income of B, C and D is Tk. 15000 per month. If the average salary of D be twice that of A, then the average salary of B and C is in Tk. 

Solution:
Given that,
Average income of A, B, C = 12,000
∴ Total income of A + B + C = 3 × 12000 = 36000 … (1)
Average income of B, C, D = 15,000
 ∴ Total income of B + C + D = 3 × 15000 = 45000 … (2)
And, D = 2A ....(3)

Now, Subtract equation (1) from (2) than we get,
B + C + D - (A + B + C) = 45000 - 36000
⇒ D - A = 9000
⇒ 2A - A = 9000   ; [from (3)]
∴ A = 9000

Now put a = 9,000 in (i)
9000 + B + C = 36000
⇒ B + C = 36000 - 9000 = 27000
∴ B + C = 27000

∴ Average salary of B and C = (B + C)/2 = 27000/2 = 13500

So the average salary of B and C is Tk. 13500.

Question: The price of a book is Tk. 200. Its price is increased by 15% and then decreased by 20%. What is the present price of the book? 

Solution: 
Initial Cost = Tk. 200
After 15% increase in the cost, it becomes,
(200 + 15% of 200)
= 200 + 30
= Tk. 230

Now, Cost is decreased by 20%, So cost will become,
(230 - 20% of 230)
= 230 - 46 
= Tk. 184

So, present cost is Tk. 184.

Question: In a class, the number of girls is 20% more than that of the boys. The strength of the class is 66. If 4 more girls are admitted to the class, the ratio of the number of boys to that of the girls is-

Solution:
Let the number of boys be 5k.
Then the number of girls = 20% more than the boys = 5k × 1.2 = 6k

Total students = 66
5k + 6k = 66
⇒ 11k = 66
⇒ k = 66/11 
∴ k = 6

∴ Number of boys = 5 × 6 = 30
∴ Number of girls = 6 × 6 = 36

Now, when 4 more girls are admitted than New number of girls = 36 + 4 = 40

∴ Ratio of boys to girls = 30 : 40 = 3 : 4 (dividing both by 10)

So the ratio of the number of boys to girls is 3 : 4

Question: 


Solution: 

সঠিক উত্তর: 3.33%
অপশনে সঠিক উত্তর না থাকায় প্রশ্নটি বাতিল করা হলো। 
---------------------- 

Question: A sum of Tk. 3600 amounts to Tk. 3960 at simple interest in 3 years. Find the rate of interest per annum.

Solution:
Principal, P = Tk. 3600
Amount after 3 years = Tk. 3960
Time, n = 3 years
Simple Interest, SI = Amount - Principal
= 3960 - 3600
= Tk. 360

We know, 
SI = (P × r × n)/100
⇒ 360 = (3600 × r × 3)/100
⇒ 360 = 36 × r × 3
⇒ r = 360/108
∴ r = 3.33%

So the rate of interest per annum is 10%.

Question: If in ΔABC, AB = 6cm, BC = 12cm and CA = 6√3cm, then the measure of ∠A is-

Solution: 

We know,
c² = a² + b²
= 6² + (6√3)² = 36 + 108 = 144 = 12² 

So, triangle ABC is a right-angled triangle at A. 
∠A = 90°

Question: The 2^nd and 8^th term of an arithmetic progression are 17 and - 1 respectively. What is the 14^th term?

Solution:
Let the first term be a and the common difference be d.

We know, 
n term of arithmetic progression = a + (n - 1)d
Then,
2^nd term, a + d = 17 ……(i)
8^th term, a + 7d = - 1 ……(ii)

Now, Subtract (i) from (ii),
(a + 7d) - (a + d) = - 1 - 17
⇒ 6d = - 18
∴ d = - 3
From (i) we  get,
⇒ a + ( - 3) = 17   ; [d =  - 3]
⇒ a - 3 = 17
∴ a = 20

Now, 14^th term = a + 13d
= 20 + 13( - 3)
= 20 - 39
= - 19

So the 14th term of the arithmetic progression is - 19.

Question: Fresh apples contain 85% water, while dry apples contain 15% water. If the weight of dry apples is 300 kg, what was the total weight of the apples when fresh?

Solution:
Given that, 
Dry apples contain 15% water
∴ they contain 85% solid matter. 
Fresh apples contain 85% water
∴ they contain 15% solid matter.

The amount of solid matter (pulp) never changes during drying.

Let the weight of fresh apples = x kg

Solid matter in fresh apples = 15% of x = 0.15x kg  
Solid matter in dry apples = 85% of 300 kg = (85/100)× 300 = 255 kg

Since the solid matter remains the same, 
0.15x = 255  
⇒ x = 255 /0.15  
⇒ x = (255 × 100)/15 
∴ x = 1700 kg

The total weight of the apples when fresh was 1700 kg.

Question: A sells an article to B at a profit of 10% B sells the article back to A at a loss of 10%. In this transaction -

Solution:
Let CP was 100 for A originally
A sells article to B at 10% profit,
CP for B = 100 + 10% of 100 = 110
Now, B sells it A again with loss 10%
Now, CP for A this time = 110 - 10% of 110 = 99
A makes Profit = 110 - 99 = 11

∴ %profit for A = (11 × 100)/100 = 11%

Question:  The supplement of an angle is twice the angle. Find the angle.

Solution:
Let the angle be x degrees.
The supplement of an angle = 180° - x

According to the question,
180° - x = 2x
⇒ 180° = 2x + x  
⇒ 180° = 3x  
⇒ x = 180°/3  
∴ x = 60°

∴ The angle is 60°

Question: 

Solution: 

Question: In a geometric progression, the 4^th term is 16 and the 7^th term is 128. Find the 10^th term.

Solution:
Let the first term = a
Common ratio = r
We know,
n-term = ar^{n - 1}

Then,
4^th term, ar³ = 16  ........(1)  
7^th term, ar⁶ = 128  ........(2)

Now, divide equation (2) by equation (1) then we get,
(ar⁶)/(ar³) = 128/16  
⇒ r³ = 8  
⇒ r³ = 2³  
∴ r = 2  

Then substitute r = 2 into equation (1) 
a (2)³ = 16   
⇒ a × 8 = 16  
∴ a = 2

Now, 10^th term
= ar⁹  
= 2 × 2⁹ 
= 2 × 2⁹  
= 2¹⁰ 
= 1024

∴ The 10th term is 1024

Question: Two buses start from a bus terminal with a speed of 60 km/h at interval of 15 minutes. What is the speed of a man coming from the opposite direction towards the bus terminal if he meets the buses at interval of 12 minutes?

Solution:
Let Speed of the man is x kmph.
Distance covered in 15 minutes at 60 kmph = distance covered in 12 minutes at (60 + x) kmph.
60 × (15/60) = (12/60)(60 + x)}
900 = 720 + 12x
12x = 180
x = 180/12
x = 15

So the speed of the man coming from the opposite direction is 15kmph 

Question: A dishonest milkman professes to sell his milk at cost price but he mixes it with water and thereby gains 25%. The percentage of water in the mixture is-

Solution:
A dishonest milkman professes to sell his milk at cost price, but he mixes it with water and gains 25%

We know, 
Profit = SP - CP
Let milk bought (cost price) be 1 litres at Tk. 100 and profit = 25

∴ Selling price = 100 + 25 = Tk. 125

As the selling price of 1 litre was Tk. 100(same as cost)

Quantity sold = 125/100 = 5/4 = 1.25 litre

Hence, water added = 1.25 - 1 = 0.25 litres

∴ Percentage of water = (0.25/1.20) × 100 = 20%

Question: A college has 8 basketball players. A 4 member's team and a captain will be selected out of these 8 players. How many different selections can be made?

Solution: 
We can select the 4 member team out of the 8 in =  ⁸C₄ ways
= 8!/4!(8 - 4)! = 8!/(4! × 4!)
= 70 ways

The captain can be selected from amongst the remaining 4 players in 4 ways.

∴ The total ways the selection of 4 players and a captain can be made = 70 × 4 ways
= 280 ways

So the total number of different selections that can be made is 280.

Question: If both the length and the breadth of a rectangle are increased by 20%, what is the percentage increase in its area?

Solution:
Let the original length = x
and breadth = y
∴ Original area = x × y = xy

After 20% increase,
 New length = x + 20% of x = x × (1 + 20/100)
= x × 1.2 = 1.2x
and new breadth = y × 1.2 = 1.2y

∴ New area = (1.2x) × (1.2y) = 1.44xy

∴ Increase in area = New area - Original area = 1.44xy - xy = 0.44xy

∴ Percentage increase in area = (Increase in area / Original area) × 100%
= (0.44xy/xy) × 100%
= 0.44 × 100%
= 44%

Question: Kamal and Rajon enter into a partnership with their capitals in the ratio 5 : 6. At the end of 6 months Kamal withdraws her capital. If they receive the profit in the ratio of 5 : 9, find how long was Rajon's capital used?

Solution: 
Let,
Kamal invested Tk. 5x for 6 months
And
Rajon invested Tk. 6x for y months
Then,
(5x × 6)/(6x × y) = 5/9
⇒ 30x/6xy = 5/9
⇒ 5/y = 5/9
∴ y = 9

So Rajon capital was used for 9 months

Question: The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is-

Solution:
The number leaves a remainder 8 when divided by 12, 15, 20 and 54.
So the required number = LCM(12, 15, 20, 54) + 8

Now, 
12 = 2 × 2 × 3
15 = 3 × 5
20 = 2 × 2 × 5
54 = 2 × 3 × 3 × 3

∴ LCM(12, 15, 20, 54) = 540

∴ Required Number = 540 + 8 = 548 

Question: A reservoir has two pipes, A and B. A can fill the reservoir 5 hours faster than B. If both together fill the reservoir in 6 hours, the reservoir will be filled by A alone in-

Solution: 
Let, A alone can fill the reservoir in x hours 
B can fill in x + 5 hours 

Both complete in 1 hour = (1/x) + (1/ x + 5)
= (2x + 5)/(x2 + 5x)

Now
1/{(2x + 5)/(x2 + 5x)} = 1/6
 (x2 + 5x)/(2x + 5) = 6 
⇒ x2 + 5x = 12x + 30 
⇒ x2 - 7x - 30 = 0
⇒ x2 - 10x + 3x - 30 = 0 
⇒ x (x - 10) + 3 (x - 10) = 0
⇒ (x - 10) (x + 3) = 0 
∴ x = 10 or, x = -3 , negative value not possible 

A alone can fill the reservoir in 10 hours

Question: The present age of Mr. Salman is three times the age of his son. Six years hence , the ratio of their ages will be 5 : 2. What is the present age of Mr. Salman?

Solution:
Let the son's age be x years ,
Then Mr. Salman's age = 3x years
Then, 
∴ (3x + 6)/(x + 6) = 5/2
⇒2(3x + 6) = 5(x + 6)
⇒ 6x + 12 = 5x + 30
⇒ 6x - 5x = 30 - 12
⇒ x = 18

∴ Present age of Mr. Salman = 3x years
= (3 × 18) years
= 54 years

Question: The area of a rectangle is equal to the area of a square with side x. If the length of the rectangle is 2x, find its breadth.

Solution:
Given that, 
A square with side x

We know,
Area of square = x²

ATQ,
Area of a rectangle is equal to the area of a square.
∴ Area of a rectangle = x2

And,
Given Length of the rectangle = 2x
Let the breadth of the rectangle = b

We know,
Area of rectangle = length × breadth
2x⋅b = x²
⇒ b = x²/2x
∴ b = x/2

Therefore, the breadth of the rectangle is x/2

Question: If √(0.0169 × x) = 1.3 than, what is the value of x?

Solution: 
Given that, 
√(0.0169 × x) = 1.3
⇒ 0.0169 × x = (1.3)²   ; [Square both sides]
⇒ 0.0169 × x = 1.69
⇒ x = 1.69/0.0169
⇒ x = (169 × 10000)/(169 × 100)
∴ x = 100

Question: The angle of elevation of a ladder leaning against a wall is 60º and the foot of the ladder is 5.5 m away from the wall. The length of the ladder is-

Solution: 

Let AB be the wall and BC be the ladder.
Then, ∠ACB = 60° = AC = 5.5 m
AC/BC = cos⁡60^∘= 1/2
⇒ BC = 2 × AC = 2 × 5.5 = 11 m

Question: Which of the following fractions is greater than 2/5 and less than 3/4 ?

Solution:
2/5 = 0.40
3/4 = 0.75

Than,   
1/3 = 0.333   
3/8 = 0.375
5/8 = 0.625
4/5 = 0.8

Clearly, 0.625 lies between 0.40 and 0.75

∴ 5/8 lies between 2/5 and 3/4.

Question: There is 60% increase in an amount in 6 years at simple interest. What will be the compound interest of Tk. 12000 after 3 years at the same rate?

Solution:
Given that,
Increase in amount after 6 years = 60%

We know,
A = P(1 + r/100)ⁿ
simple Interest = (principal × rate × time)/100

Now,
Let the amount at 1^st year be 100x
∴ Increased in amount = 60% of 100x = 60x
⇒ 60x = (100x × 6 × r)/100
⇒ 6r = 60
⇒ r = 60/6 = 10
∴ r = 10% 

∴ Compound Interest after 3 years = P(1 + r/100)ⁿ - P
= 12000(1 + 10/100)³ - 12000
= 12000 × (11/10)³ - 12000
= 12000 × (11/10) × (11/10) × (11/10) - 12000
= 15972 - 12000
= Tk. 3972

∴ The required answer is Tk. 3972

Question: The ratio between the perimeter and the length of a rectangle is 3 : 1. If the area of the rectangle is 50 sq. cm, what is the breadth of the rectangle?

Solution:
Let the length and breadth be x and y respectively

So, 2(x + y) : x = 3 : 1
⇒ (2x + 2y)/x = 3/1
⇒ 2x + 2y = 3x
⇒ 2y = 3x - 2x
⇒ 2y = x
∴ x = 2y

Then,
Area = 50
⇒ x × y = 50
⇒ 2y × y = 50
⇒ 2y² = 50
⇒ y² = 25 = 5²
∴ y = 5

So the breadth of the rectangle is 5 cm.

Question: If P and Q together can complete a work in 18 days, P and R together in 12 days, and Q and R together in 9 days, then Q alone can do the work in-

Solution:
One day's work of,
(P + Q) = 1/18 .......(1)

One day's work of,
(P + R) = 1/12 .......(2)

One day's work of,
(Q + R) = 9 .......(3)

Adding(1),(2)and(3),
⇒ 2 × (P + Q + R) = (1/18) + (1/12) + (1/9)
⇒ 2 × (P + Q + R) = 1/4
⇒ (P + Q + R) = 1/8

Now,
⇒ Q = (1/8) - (P + R)
⇒ Q = (1/8) - (1/12)
One day's work of Q = (3 - 2)/24 = 1/24
∴ Q need 24 days

Question: A bag contains 6 red, 5 blue, and 4 green balls. Two balls are drawn one after another without replacement. What is the probability that both balls are red?

Solution:
Total balls = 6 + 5 + 4 = 15
Probability of first red = 6/15
Probability of second red = 5/14

∴ Probability (both red) = (6/15) × (5/14)
= 30/210
= 1/7

Therefore, the probability that both balls are red is 1/7.

Question: If a is an even integer and b is an odd integer, which of the following must be an odd integer?

Solution: 
Let, a = 2 and b = 3

Then, 
ক) a/b
= 2/3
So a/b is not an integer, hence cannot be odd integer.

খ) 2(a + b)
= 2 × (2 + 3)
= 10
So 2(a + b) is always even.

গ) ab
= 2 × 3
= 6    ; [even × odd = even]
So ab is always even.

ঘ) 2a + b 
= (2 × 2) + 3
= 4 + 3
= 7
So 2a + b is always odd.

∴ 2a + b This must be an odd integer.

Question: A hall is 30m long and 10m broad. If the sum of the areas of the floor and the ceiling is equal to of the areas of four walls, the volume of the hall is-

Solution:
Let,
The height of the hall = h

Given that, 
Length of the room = 30 m
Breadth of the room = 10 m

According to the question,
2 × (30 × 10) = 2 × (30 + 10) × h
⇒ 2 × 40 × h = 2 × (30 × 10)
⇒ 40h = 300
⇒ h = 300/40
⇒ h = 30/4
∴ h = 15/2

We know,
 Volume = 30 × 10 × (15/2)
= 2250 m³

Question: Which of the following is equivalent to the pair of inequalities 2x - 5 ≤ 7 and 3x + 4 > 10?

Solution:
Solve the first inequality,
2x - 5 ≤ 7 
⇒ 2x ≤ 7 + 5
⇒ 2x ≤ 12
∴ x ≤ 6
And,
Solve the second inequality,
3x + 4 > 10 
⇒ 3x > 10 - 4
⇒ 3x > 6
∴ x > 2

∴ We get 2 < x ≤ 6

Question: In a tourist group of 100 people, 55 speak French, 40 speak Spanish, and 20 speak none of the languages. How many of them speak just one language?

Solution:

Let,
Number of people who can speak both languages = x persons
∴ Number of people who speak only French = (55 - x) persons
∴ Number of people who speak only Spanish = (40 - x) persons

Given that,
Number of people who speak none of the languages = 20 persons

According to the question,
Only French + Both + Only Spanish = Total students - Those who speak none
⇒ (55 - x) + x + (40 - x) = 100 - 20 
⇒ 95 - x = 80
⇒ x = 95 - 80
∴ x = 15

∴ Only French = (55 - 15) = 40 persons
∴ Only Spanish = (40 - 15) = 25 persons

∴ Number of people who speak only one language (French or Spanish) = (40 + 25) = 65 persons

Question: A tank is 1/3 parts full with water. If 9 liters of water is added, the tank becomes 5/6 parts full. What is the capacity of the tank?

Solution:
Let the capacity of the tank = x liters

According to the question,
(x/3) + 9 = 5x/6
⇒ (5x/6) - (x/3) = 9
⇒ (5x - 2x)/6 = 9
⇒ 3x = 54
⇒ x = 54/3
⇒ x = 18

Therefore, the capacity of the tank = 18 liters. 

Question: If (x/y) + (y/x) = √8 then what is the value of (x⁴/y⁴) + (y⁴/x⁴) ?

Solution:
Given that, 
(x/y) + (y/x) = √8

∴ x⁴/y⁴ + y⁴/x⁴
= (x/y)⁴ + (y/x)⁴
= {(x/y)²}² + {(y/x)²}²
= {(x/y)² + (y/x)²}² - 2.(x²/y²).(y²/x²)
= {(x/y)² + (y/x)²}² - 2
= [{(x/y) + (y/x)}² - 2.(x/y).(y/x)]² - 2
= {(√8)² - 2}² - 2
= (8 - 2)² - 2
= 6² - 2
= 36 - 2
= 34

Question: If ABC and PQR are similar triangles in which ∠A = 47° and ∠Q = 83°, then ∠C is-


Solution: 
Since, ΔABC and ΔPQR are similar triangles.
And ∠A = 47°
then,
∠B = ∠Q = 83°

Thus, in ΔABC,
∠A + ∠B + ∠C = 180°
∠C = 180° - (∠A + ∠ B)
⇒ ∠C = 180° - (47° + 83°)
∴ ∠C = 50°

Question: What is the slope of a line perpendicular to the line whose equation is 20x - 2y = 6?

Solution:
The general equation of a straight line is
y = mx + c ......(1) (Where, m = slope)

If the slope of a line is m, then the slope of the line perpendicular to it is,
m' = - (1/m)

Now,
20x - 2y = 6
⇒ 2y = 20x - 6
∴ y = 10x - 3
Comparing with equation (1), we get,
∴ m = 10

∴ The slope of the perpendicular line is, m' = - (1/10)

Question: A man swimming in a stream which flows (3/2) km/hr finds that in a given time he can swim twice as far with the stream as he can against it. At what rate does he swim?

Solution:
Let the rate of his swim x km/h
When he swim with the flow then speed =(x + 3/2) km/h
∴ S₁ = (x + 3/2) × t
When he swim against the flow stream then speed = (x - 3/2) km/h
∴ S₂ = (x - 3/2) × t

According to the question,
S₁ = 2 × S₂
⇒ (x + 3/2)t = 2(x - 3/2)t
⇒ (2x + 3)/2 = 2x - 3
⇒ 2x + 3 = 4x - 6
⇒ 9 = 2x
⇒ x = 9/2
∴ x = 4.5 km/hr

Therefore, the man swims at 9/2 km/h (or 4.5 km/h) in still water.

Question: If an exponent or index has base 15 and power zero, then which of the following will be its value?

Solution:
a⁰ = 1 (for any non-zero base a)

Now,
= 15⁰
= 1

Question: There are two squares S₁ and S₂. The ratio of their areas is 9 : 16. If the side of S₁ is 12 cm. What is the side of S₂?

Solution:
Given that,
Two squares S_1​ and S₂​
Area of S₁ : Area of S₂ = 9 : 16
Side of S₁ = 12 cm

Now,
Let the side of S_2​ be x cm.
Then,
(Side of S₁)² : (Side of S₂)² = 9 : 16
⇒ 12² : x² = 9 : 16
⇒ 144/x² = 9/16
⇒ x² = (144 × 16)/9
⇒ x² = 256 
⇒ x² = 16²
∴ x = 16

So the side of S₂​ is 16 cm.