পেট্রোবাংলা প্রিলি ও লিখিত সমন্বিত প্রস্তুতি [Archived]

Question: A man can do a work in 20 days and a woman in 15 days. If they work on it together for 6 days, then the fraction of the work that is left is-

Solution:
Man’s 1 day’s work = 1/20
Woman’s 1 day’s work = 1/15

(Man + woman)’s 1 day’s work = (1/20 + 1/15)  = 7/60
(Man + woman)’s 6 day’s work = (7/60 × 6) = 7/10

Thus, Remaining work = 1 - 7/10 = 3/10

Question: Solve the inequality: ∣x - 3∣ ≥ 4

Solution:
Consider two cases:
(i) x - 3 ≥ 4
Add 3 to both sides: x ≥ 7

(ii) x - 3 ≤ - 4
Add 3 to both sides:
x ≤ - 1

So, the solution to ∣x - 3∣ ≥ 4 is x ≤ - 1 or x ≥ 7

Question: Raju hikes up a hill at 4 mph and comes down at 6 mph. If the total time taken for the total journey is 3.5 hours, what was the distance between the hilltop and the foothills?

Solution:
Average speed = 2ab/(a + b) = (2 × 6 × 4)/10 = 4.8 mph.
Time taken = 3.5 hours both ways.

So, the two way distance = 4.8 × 3.5 miles = 16.8 miles.

Hence, the distance one-way = 16.8/2 = 8.4 miles.

Question: Two pipes A and B can fill a tank in 20 hours and 30 hours respectively. If both the pipes are opened simultaneously, find after how much time should pipe B be closed so that the tank is full in 18 hours?

Solution:
Let the capacity of the tank be LCM (20, 30) = 60 units
Efficiency of pipe A = 60/20 = 3 units/hour
Efficiency of pipe B = 60/30 = 2 units/hour 
Combined efficiency of pipes A and B = 5 units/hour

Let both A and B be opened for ‘n’ hours and then, B be closed and only A be opened for the remaining ’18 - n’ hours.
5n + 3 × (18 - n) = 60
⇒ 2n + 54 = 60 
⇒ 2n = 6
∴ n = 3
Therefore, B should be closed after 3 hours.

Question: The area of a rectangle is four times of a square. The length of the rectangle is 80 cm and the breadth of the rectangle is 3 times that of the side of the square. What is the side of the square?

Solution:
L = 80 cm.
B = 3a, where a is the side of the square.

ATQ,
Area of rectangle = LB = 4a²
 ⇒ 80 × 3a = 4a²
⇒ 240 = 4a
∴ a = 60 cm.

Question: 3 pumps, working 4 hours a day, can empty a tank in 2 days. How many hours a day must 4 pumps work, to empty the tank in one day?

Solution:
As number of pumps increase, the time required decreases and when working hours increase, fewer days are required to complete the work. Hence, this is a problem related to indirect proportion.

More pumps (↑),Less working hours (↓)
More working hours (↑),Less days (↓)


⇒ 4 × 3 × 2 = 4 × 1 × x
⇒ 24 = 4x
∴ x = 6

Question: A man travelled a distance of 61 km in 9 hours. He travelled partly on foot at 4 km/hr and partly on bicycle at 9 km/hr. What is the distance travelled on foot?

Solution:
Let the time in which he travelled on foot = x hr
Then the time in which he travelled on bicycle =(9 - x) hr
distance = speed × time
⇒ 4x + 9(9 - x) = 61
⇒ 4x + 81 - 9x = 61
⇒ 5x = 20
⇒ x = 4

∴ The distance travelled on foot = 4 × 4 = 16 km

Question: Time required by two pipes A and B working separately to fill a tank is 36 seconds and 45 seconds respectively. Another pipe C can empty the tank in 30 seconds. Initially, A and B are opened and after 7 seconds, C is also opened. In how much more time the tank would be completely filled?

Solution:
Let the capacity of the tank be LCM (36, 45, 30) = 180 units 
Efficiency of pipe A = 180 / 36 = 5 units/second 
Efficiency of pipe B = 180 / 45 = 4 units/second 
Efficiency of pipe C = - 180 / 30 = - 6 units/second

Now, for the first 7 seconds, A and B were open. 
Combined efficiency of A and B = 5 + 4 = 9 units/second 
Part of the tank filled in 7 seconds = 7 × 9 = 63 units 

Part of tank empty = 180 - 63 = 117 units

Now, all pipes are opened. 
Combined efficiency of all pipes = 5 + 4 - 6 = 3 units/second

Therefore, more time required = 117/3 = 39 seconds.

Question: A train of length 240 meters crosses a pole in 12 seconds. In what time it will cross a platform of length 400 meters?

Solution:
First, we need to find the speed of the train when it crosses the pole.
Length of the train: 240 meters
Time to cross the pole: 12 seconds
The speed of the train (S) can be calculated using the formula:
S = Distance/Time = 240 meters/12 seconds = 20 m/s

When crossing the platform, the train needs to cover its own length plus the length of the platform.
Length of the platform: 400 meters
Total distance to cross: Length of the train + Length of the platform
Total distance = 240 meters + 400 meters = 640 meters

∴ Required time = 640/20 seconds = 32 seconds

Question: Hemal completes a job in 45/2 days. What part of the job will he do in 2 days?

Solution:
We know, if a person does a job in n days, then his 1-day work = 1/n
Here,
n = 45/2
Hemal’s 1-day work = 2/45
Thus, Hemal’s 2 days work = 2 × (2/45) = 4/45

Question: A wheel that has 6 cogs is meshed with a larger wheel of 12 cogs. If the smaller wheel has made 22 revolutions, then find the number of revolutions made by the larger wheel.

Solution:
As number of cogs increase, the revolutions made decrease. Hence, this is a problem related to indirect proportion.
Let the number of wheels be x.
More cogs (↑),Less revolutions (↓)

12 : 6 : : 22 : x
⇒ 12 × x = 6 × 22
⇒ x = (6 × 22)/12
∴ x = 11

Question:  Walking 6/7th of his usual speed, a man is 12 minutes too late. What is the usual time taken by him to cover that distance?

Solution:
New speed =6/7 of usual speed
Speed and time are inversely proportional.
Hence new time = 7/6 of usual time
Hence, 
7/6 of usual time - usual time = 12 minutes
⇒ 1/6 of usual time = 12 minutes
∴ usual time = 12 × 6 = 72 minutes = 1 hour 12 minutes

Question: Two small circular parks of diameters 6 m and 8 m are to be replaced by a bigger circular park. What would be the radius of this new park, in meter, if the new park occupies the same space as the two small parks (in meter)?

Solution:
Let,
The radious of the new circular park = R
Area of the new circular park = sum of the areas of the 2 smaller parks
⇒ π (6/2)² + π (8/2)² 
= π (3)² + π (4)²
= π 9 + π 16
= π(9 + 16)
= 25π

⇒ 25 π = π R².
∴ R² = 25
⇒ R = 5 m

Question: A man rows 24 km upstream in 6 hours and a distance of 35 km downstream in 7 hours. Then the speed of the man in still water is

Solution:
Speed of upstream = 24/6 = 4 km/hr.
Speed of downstream = 35/7 = 5km/hr.

∴ Speed of man in still water = (4 + 5)/2 = 4.5 km/hr.

Question: If 30 men can build a wall 56 meters long in 5 days, what length of a similar wall can be built by 40 men in 3 days?

Solution:
 If more men work, length of wall built is more. If worked for few days, the length of wall built is also less. Hence, this problem is related to direct proportion.
The two main parameters are man and days.
Therefore,

⇒ 30 × 5 × x = 40 × 3 × 56
⇒ x = (40 × 3 × 56)/(30 × 5)
∴ x = 44.8

Question: The speed of a car increases by 2 kms after every one hour. If the distance travelling in the first one hour was 35 kms. what was the total distance travelled in 12 hours?

Solution:
Total distance travelled in 12 hours = (35 + 37 + 39 + ..... upto 12 terms)
This is an A.P with first term, a = 35,
number of terms, n= 12,
d=2.

Required distance  = (12/2)[2 × 35 + {12 - 1) × 2]
= 6(70 + 22)
= 552 kms

Question: Three pipes A, B and C were opened to fill a cistern. Working alone, A, B and C require 12, 15 and 20 minutes respectively. Another pipe D, which is a waste pipe, can empty the filled tank in 30 minutes working alone. What is the total time (in minutes) taken to fill the cistern if all the pipes are simultaneously opened?

Solution:
Let the capacity of the cistern be LCM(12, 15, 20, 30) = 60 units.
Efficiency of pipe A = 60 / 12 = 5 units / minute
Efficiency of pipe B = 60 / 15 = 4 units / minute
Efficiency of pipe C = 60 / 20 = 3 units / minute
Efficiency of pipe D = 60 / 30 = 2 units / minute

Combined efficiency of pipe A, pipe B, pipe C and pipe D = 10 units/minute  
Therefore, time required to fill the cistern if all the pipes are opened simultaneously = 60/10 = 6 minutes

Question: Two trains of equal length are running on parallel lines in the same direction at 46 km/hr and 36 km/hr. The faster train passes the slower train in 36 seconds. The length of each train is-

Solution:
Let the length of each train be x metres.
Then, distance covered = 2x metres.
Relative speed = (46 - 36) km/hr
= 10(5/18) m/sec
= 25/9  m/sec

ATQ,
2x/36 = 25/9
⇒ 2x =100
∴ x = 50 m

Question: P, Q and R together can complete a work in 16 days and R alone complete the work in 20 days. If P, Q and R started the work together and after 10 days P and Q left the work, in how many days R alone complete the remaining work?

Solution:
P + Q + R = 16 days
R =20 days

work (LCM of 16 and 20) = 80
( P + Q +R) ‘s work 
= 80 /16
= 5 unit

Work done by R = 80 /20 = 4 unit

(P + Q + R) ‘s 10 days work 
= 5 × 10 
= 50 unit

Remaining work 
= (80 - 50) 
= 30 unit

Remaining work done by R 
= 30/4 
= 7.5 days

Question: For sets A = {x | x is an integer, 1 ≤ x ≤ 6} and B = {x | x is an even integer, 2 ≤ x ≤ 8}, find the set A - B.

Solution:
A = {x | x is an integer, 1 ≤ x ≤ 6} 
A = {1, 2, 3, 4, 5, 6},

B = {x | x is an even integer, 2 ≤ x ≤ 8},
B = {2, 4, 6, 8}.

A - B (difference) is the set of elements in A that are not in B.
A - B = {1, 3, 5}.

Question: A thief is noticed by a policeman from a distance of 200 m. The thief starts running and the policeman chases him. The thief and the policeman run at the rate of 10 km and 11 km per hour respectively. What is the distance between them after 6 minutes?

Solution:
Relative speed of the thief and policeman  =  (11 - 10) km/hr = 1 km/hr 
Distance covered in 6 minutes  = (1/60) × 6 km   = 1/10 km = 100 m
Therefore, Distance between the thief and policeman = (200 - 100) m = 100 m.

Question: Working alone, two pipes A and B require 9 hours and 6.25 hours more respectively to fill a pool than if they were working together. Find the total time taken to fill the pool if both were working together.

Solution:
Let the time taken if both were working together be 'n' hours.
Time taken by A = n + 9
Time taken by B = n + 6.25  

In such kind of problems, we apply the formula :
n² = a × b, where 'a' and 'b' are the extra time taken if both work individually than if both work together.
Therefore,
n² = 9 × 6.25
⇒ n = 3 × 2.5 = 7.5  

Thus, working together, pipes A and B require 7.5 hours

Question: A man can row upstream 10 kmph and downstream 20 kmph. Find the man rate in still water and rate of the stream.

Solution:
If a is rate downstream and b is rate upstream 
Rate in still water = (a + b)/2 
Rate of current = (a - b)/2 

Rate in still water = (20 + 10)/2 = 15 kmph 
Rate of current = (20 - 10)/2 = 5 kmph 

Question: M did a piece of work in 5 days. That piece of work was done by N in 9 days. If M and N worked together, they got total wages of Tk. 4200. Find the share of N.

Solution:
M's 1 day's work 1/5
N's 1 day's work 1/9


M : N
Time = 5 : 9
Efficiency = 9 : 5

(Time and efficiency are inversely proportional) 
N gets = 4200 ×  (5/14)
= 1500

Thus, N gets the wages of Tk. 1500.

Question: If 8 men can reap 40 hectares in 12 days, then how many hectares can 30 men reap in 20 days?

Solution:
Let the required number of hectares be x.

More men (↑) More hectares (↑)

⇒ 8 × 12 × x = 30 × 20 × 40
⇒ x = (30 × 20 × 40)/(8 × 12)
∴ x = 250

Question: In how many ways can a group of 5 members be formed by selecting 3 boys out of 6 boys and 2 girls out of 5 girls?

Solution:
Number of ways 3 boys can be selected out of 6 = ⁶C₃ = 6!/[(3!) × (3!)] = (6 × 5 × 4) / (3 × 2 × 1) = 20

Number of ways 2 girls can be selected out of 5 = ⁵C₂ = 5!/[(2 !) × (3 !)] = (5 × 4)/(2 × 1) = 10

Therefore, total number of ways of forming the group = 20 × 10 = 200 

Question: log(a/b) + log(b/c) + log(c/a) = ?

Solution:
log(a/b) + log(b/c) + log(c/a)
= loga - logb + logb - logc + logc - loga
= 0

Question: If x is an integer then solve (log₂ x)² - log₂ x⁴ - 32 = 0.

Solution:
We have (log₂x)² - log₂x⁴ - 32 = 0.
⇒ (log₂x)² - 4log₂x - 32 = 0 ......(1)
Let log₂x = y
(i) ⇒ y² - 4y - 32 = 0
⇒ y² - 8y + 4y - 32 = 0
⇒ y(y - 8) + 4(y - 8) = 0
⇒ (y - 8)(y + 4) = 0
⇒ y = 8, - 4
⇒ log₂x = 8 or log₂x = - 4
⇒ x = 2⁸ = 256 or x = 2^{- 4} = 1/16
Since ‘x’ is an integer so x = 256.

Question: Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?

Solution:
Here, S = {1, 2, 3, 4, ...., 19, 20}.

Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.

P(E) = n(E)/n(S) = 9/20.

Question: How many words can be formed by using the letters from the word 'DRIVER' such that all the vowels are always together?

Solution:
In these types of questions, we assume all the vowels to be a single character, i.e., “IE” is a single character.

So, now we have 5 characters in the word, namely, D, R, V, R, and IE. But, R occurs 2 times.

 Number of possible arrangements = 5!/2! = 60
Now, the two vowels can be arranged in 2! = 2 ways.

Total number of possible words such that the vowels are always together = 60 × 2 = 120 

Question: In a college, 200 students are randomly selected. 140 like tea, 120 like coffee and 80 like both tea and coffee. How many students like only tea?

Solution:
The given information may be represented by the following Venn diagram, where T = tea and C = coffee.


Number of students who like only tea = 140 - 80 = 60