পেট্রোবাংলা প্রিলি ও লিখিত সমন্বিত প্রস্তুতি [Archived]

Question: In a race of 1000 m, A can beat B by 100 m. In a 400 m, B beats C by 40 m. In a race of 500 m. A will beat C by:

Solution:
When A runs 1000 m, B runs 900 m.
Hence, when A runs 500 m, B runs 450 m.
Again, when B runs 400 m, C runs 360 m.

And, when B runs 450 m, C runs = 450 × (360/400) = 405 m.

∴ Required distance = 500 - 405 = 95 m.

Hence, In a race of 500 m. A will beat C by 95 m.

Question: A tank is filled in 5 hours by three pipes A, B and C. The pipe C is twice as fast as B and B is twice as fast as A. How much time will pipe A alone take to fill the tank?

Solution:
Suppose,
Pipe A alone takes x hours to fill the tank.
So, pipes B and C will take x/2 and x/4 hours respectively to fill the tank.

Now,
1/x + 2/x + 4/x = 1/5
⇒ (1 + 2 + 4)/x = 1/5
⇒ 7/x = 1/5
∴ x = 35

∴ Pipe A alone takes 35 hours to fill the tank.

Question: A monkey climbs a 60 m high pole. In the first minute, he climbs 6 m and slips down 3 m in the next minute. How much time is required by it to reach the top?

Solution:
According to the question,
For 1^st min, he climbs 6 m and comes down 3 m in 2^nd min.
Effectively in 2 min, he climbs 3 m.

In 1 min, he climbs (3/2) m

So, in 36 mins he climbs = (3/2) × 36 = 54 m

So, in the 37^th minute, he climbs the next 6 m.

∴ The required value is 37 minutes.

Question: 3/4 part of the tank is full of water. When 30 litres of water is taken out, the tank becomes empty. The capacity of the tank is -

Solution:
Let us consider,
The tank has 4x litres of total capacity and holds 3x litres of water.
And if 30 litres of water is taken out, then the tank becomes empty.

It means 3x litres of water is taken out.
∴ 3x = 30 litres
⇒ x = 10 litres

Capacity of tank
= 4x
= 4 × 10
= 40 liters.

Question: A man takes 6 hours 15 minutes in walking a distance and riding back to starting place. He could walk both ways in 7 hours 45 minutes. The time taken by him to ride back both ways is:

Solution:
Time taken in walking both the ways = 7 hours 45 minutes - - - - - - - - (1)
Time taken in walking one way and riding back = 6 hours 15 minutes - - - - - (2)

By the equation (2) × 2 - (1), we have,
Time taken by the man in riding both ways = 12 hours 30 minutes - 7 hours 45 minutes
= 4 hours 45 minutes
= 285 minutes

Question: 12 buckets of water fill a tank when the capacity of each bucket is 13.5 litres. How many buckets will be needed to fill the same tank, if the capacity of each bucket is 9 litres?

Solution:
Capacity of the tank
= (12 × 13.5) litres
= 162 litres

Capacity of each bucket = 9 litres

∴ Number of buckets needed
= 162/9
= 18

Question: The ratio of speeds of A and B is 2 : 3 and therefore A takes 20 minutes more time than B. What is the ratio of time taken by A and B?

Solution:
When distance is constant then speed is inversely proportional to time.
ST = D

When distance is constant,
S ∝ 1/T

So, the ratio of time taken by A and B = 3 : 2

Question: Tap A can fill a tank in 6 hours, tap B can fill the same tank in 8 hours and tap C can empty the same tank in 4 hours. If all three taps A, B and C are opened together, then how much time (in hours) will be taken to fill the tank?

Solution:
Total unit of work = 24 units [LCM of 6 hours, 8 hours and 4 hours]

Efficiency of A = 24/6 = 4
Efficiency of A = 24/8 = 3
Efficiency of A = 24/4 = 6

Total work in a hours by (A + B + C) = 4 + 3 - 6 = 1 unit

Hence, time will be taken to fill the tank = 24/1 = 24 hours

Question: A dog starts chasing to a cat 2 hours later. It takes 2 hours to dog to catch the cat. If the speed of the dog is 30 km/h, what is the speed of cat?

Solution:
Let the speed of the cat be = x km/hr
The speed of the dog be = 30 km/hr

Distance covered by cat in 2 hrs = (x × 2) = 2x km

Relative speed = (30 - x) km/hr

t = s/v
⇒ 2 = 2x/(30 - x) [It takes 2 hours to dog to catch the cat, so t = 2]
⇒ x = 30 - x
⇒ 2x = 30
∴ x = 15 km/hr

Question: A man completes a certain journey by a car. If he covered 30% of the distance at the speed of 20kmph. 60% of the distance at 40km/h and the remaining of the distance at 10 kmph, his average speed is:

Solution:
Suppose, total distance = 100 km

He covered 30% of the distance at the speed of 20kmph
His time taken = 30/20 hours

He covered 60% of the distance at 40km/h
His time taken = 60/40 hours

He covered the remaining of the distance at 10 kmph
The remaining of the distance = 100% - (30 + 60)% = 10%
His time taken = 10/10 hour

∴ Average speed = total distance/time
= 100/(30/20 + 60/40 + 10/10)
= 25 km/h

Question: A water tap fills a tub in 'p' hours and a sink at the bottom empties it in 'q' hours. If p < q and both tap and sink are opened the tank is filled in 'r' hours, then the relation between p, q, r :

Solution:
Total unit of work = pq unit

The efficiency of the water tap that fills a tub = pq/p = q
The efficiency of the sink at the bottom that empties the tub = pq/q = p

Net efficiency = q - p [q > p]

Hence, Time required to fill the tub, r = pq/(q - p)
⇒ 1/r = (1/p) - (1/q)

Question: An outlet pipe can empty a cistern in 3 hours. In what time will empty 2/3 of the cistern?

Solution:
The outlet pipe empties the one complete cistern in 3 hours

∴ Time taken to empty 2/3 of the cistern
= (2/3) × 3
= 2 hours

Question: A tiger is 50 of its own leaps behind a deer. The tiger takes 5 leaps per minute to the deer's 4. If the tiger and the deer cover 8 m and 5 m per leap respectively, what distance will the tiger have to run before it caches the deer?

Solution:
Speed of tiger = 40 m/min
Speed of deer = 20 m/min.

Relative speed = 40 - 20 = 20 m/min.
Initial difference in distance = 50 × 8 = 400 m

Time taken to catch = 400/20
 = 20 min.

Distance travelled in 20 min,
= 20 × 40
= 800 m

Question: 6 pipes, working 10 hours a day, can empty a cistern in 3 days. How many hours a day must 9 pipes work to empty the cistern in one day?

Solution:
By applying the MDH method,
it can be written as,

6 × 10 × 3 = 9 × x × 1
⇒ x = 20 hours

Question: In a race of 200 meters, B can give a start of 10 meters to A, and C can give a start of 20 meters to B. The starts that C can give to A, in the same race is:

Solution:
According to the question,
When B runs 200 m meters, A runs 190 meters;
Hence, when B runs 180 meters,

A runs = 190 × (180/200) meters
= 171 meters

When C runs 200m, B runs 180 meters.

Hence,
C will give a start to A by = (200 - 171) meters
= 29 meters

Question: The volume of a tank is 72 cubic metres. Water is poured into it at the rate of 60 litres per minute. How much time will it take to fill the tank?

Solution:
Time will be taken to fill the tank
= (7200/60) minutes
= 1200 minutes
= (1200/60) hours
= 20 hours

Question: An athlete runs 200 meters race in 24 seconds. His speed in km/h is-

Solution:
Speed of athlete
= 200 m/24 sec
= (200 × 18)/(5 × 24)
= 30 km/hour

Question: A cyclist moving on a circular track of a radius of 100 meters completes one revolution in 2 minutes. What is the average speed of a cyclist (approx.)?

Solution:
Distance covered in one complete revolution
= 2πr
= 2 × 100 × (22/7)
= 628.571 meters
≈ 628 meters

Hence,
Average speed 
= 628/2
= 314 meters/minute

Question: A pipe can fill a tank in x hours and another pipe can empty it in y (y > x) hours. If both pipes are open, in how many hours will the tank is filled?

Solution:
Net part filled in 1 hour
= (1/x) - (1/y)
= (y - x)/xy hours

∴ The tank will be filled in = xy/(y - x) hours.

Question: A pump can fill a tank with water in 2 hours. Because of a leak, it took 2 hours and 20 minutes to fill the tank. The leak can drain all the water of the tank in:

Solution:
2 hours and 20 minutes = 7/3 hours [7/3 hours = (7/3) × 60 = 140 minutes]

Work done by the leak in 1 hour 
= (1/2) - (3/7)
= (7 - 6)/14
= 1/14

∴ Leak will empty the tank in 14 hours

Question: The speed of A and B are in the ratio 3 : 4. A takes 30 minutes more than B to reach a destination. Time in which A reach the destination?

Solution:
Ratio of speed = 3 : 4
Ratio of time taken = 4 : 3 (As Speed ∝ 1/Time, When distance remains constant)

Let the time taken by A and B be 4x and 3x hours respectively.

Then,
4x - 3x =  30/60
Or, x = 1/2

Hence, time taken by A = 4x = 4 × (1/2) = 2 hours = 120 minutes

Question: A man reduces his speed from 20 kmph to 18 kmph. So, he takes 10 minutes more than the normal time. what is the distance travelled by him?

Solution:
As the speed decreases from 20 kmph to 18 kmph i.e. 10 % increment in usual time.
10% = 10 min.
100% = 100 min.

Now,
Distance travelled by him,
= (100/60) × 18
= 30 km

Question: A tap can fill a tank in 6 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the tank completely?

Solution:
Time taken by one tap to fill half of the tank = 3 hours

Partly filling the tank with the four taps in 1 hour = 4 × (1/6) = 2/3

Remaining part = 1 - (1/2) = 1/2

∴ 2/3 : 1/2 :: 1 : x
⇒ x = (1/2) × 1 × (3/2)
⇒ x = 3/4 hours
⇒ x = 45 minutes

Hence, the total time taken = 3 hours 45 minutes

Question: An individual is cycling at a speed of 25 km per hour. He catches his predecessor who had started earlier in two hours. What is the speed of his predecessor who had started 3 hours earlier?

Solution: 
The distance covered in two hours,
= 2 × 25 = 50 km

Time taken by Predecessor = (3 + 2) = 5 hours

Then, the speed of the predecessor,
= 50/5
= 10 kmph

উত্তর None of the above হবে, কারণ speed এর একক kmph.

Question: One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank in:

Solution:
Let the slower pipe alone fill the tank in x minutes.

Then, the faster pipe will fill it in x/3 minutes.

∴ (1/x) + (3/x) = 1/36
⇒ 4/x = 1/36
⇒ x = 144 minutes
∴ x = 2 hours 24 minutes

Question: A certain distance is covered at a certain speed. If half the distance is covered in double the time, the ratio of the two speeds is:

Solution:
Let x km be covered in y hours.

Then, speed = (x/y) km/hr

In the second case, x/2 km is covered in 2y hours.

∴ New speed = (x/2) × (1/2y ) km/hr = (x/4y) km/hr

∴ Ratio of speeds = (x/y) : (x/4y)
= 1 : (1/4)
= 4 : 1

Question: Two pipes can fill a tank in 20 and 24 minutes respectively and a waste pipe can empty 3 gallons per minute. All three pipes working together can fill the tank in 15 minutes. The capacity of the tank is:

Solution:
Let the waste pipe empty the tank in x minutes.

According to the question,
(1/20 + 1/24) - 1/x = 1/15
⇒ 1/x = (1/20 + 1/24) - 1/15
⇒ 1/x = 1/40
∴ x = 40 minutes

A waste pipe can empty 3 gallons per minute
In 40 minutes it can empty = 3 × 40 = 120 gallons.

∴ Capacity of the tank = 120 gallons.

Question: A is twice as fast as B and B is thrice as fast as C. The journey covered by C in 42 minutes will be covered by A in:

Solution:
Let C's speed be x metres/min
Let the time taken by A be y min

Then, B's speed = 3x metres/min
And, A's speed = 6x metres/min

Ratio of speed of A and C = Ratio of time taken by C and A

6x : x = 42 : y
⇒ 6x/x = 42/y
∴ y = 7 minutes

Hence, The journey covered by C in 42 minutes will be covered by A in 7 minutes.

Question: Pipes A and B can fill a tank in 5 and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in:

Solution:
Net part filled in 1 hour = (1/5) + (1/6) - (1/12)
= {(12 + 10) - 5}/60
= 17/60

Hence, the tank will be filled in 60/17 hours.