First find the distance to the country house.

Using \(s=vt\):

\(s=60\cdot1.5=90\text{ km}\).

On the way back, Akhror traveled the same distance \(90\text{ km}\) at \(45\text{ km/h}\).

So the return time is:

\(t=\dfrac{s}{v}=\dfrac{90}{45}=2\text{ h}\).

Let the second administrator service \(x\) computers per hour.

Then the first services \(x+2\) computers per hour.

Their total rate is:

\(x+(x+2)=2x+2\).

In 4 hours they serviced 48 computers, so

\(4(2x+2)=48\).

\(2x+2=12\).

\(2x=10\).

\(x=5\).

Therefore, the second administrator services 5 computers per hour, and the first services \(5+2=7\).

Let the first line assemble \(x\) items per hour.

Then the second assembles \(x+3\) items per hour.

Their total rate is:

\(x+(x+3)=2x+3\).

In 5 hours they assembled 155 items, so

\(5(2x+3)=155\).

\(2x+3=31\).

\(2x=28\).

\(x=14\).

Therefore, the first line assembles 14 items per hour, and the second line 17 items per hour.

Let the second team repair \(x\) meters per day.

Then the first team repairs \(x-10\) meters per day.

In one day together they repair:

\(x+(x-10)=2x-10\) meters.

In 4 days:

\(4(2x-10)=840\).

\(2x-10=210\).

\(2x=220\).

\(x=110\).

Therefore, the second team repairs 110 m per day, and the first team 100 m per day.

Let the second crane unload \(x\) containers per hour.

Then the first crane unloads \(x+5\) containers per hour.

Their total rate is:

\(x+(x+5)=2x+5\).

In 4 hours they unloaded 180 containers, so

\(4(2x+5)=180\).

\(2x+5=45\).

\(2x=40\).

\(x=20\).

Therefore, the second crane unloads 20 containers per hour, and the first unloads 25 containers per hour.

Find the distance between the cities:

\(s=75\cdot4=300\text{ km}\).

On the way back, the motorcyclist traveled for 5 hours.

His return speed was:

\(v=\dfrac{300}{5}=60\text{ km/h}\).

The difference in speeds is:

\(75-60=15\text{ km/h}\).

Find the total distance.

In the first 2 hours, the car traveled:

\(80\cdot2=160\text{ km}\).

In the next 4 hours:

\(50\cdot4=200\text{ km}\).

Total distance:

\(160+200=360\text{ km}\).

Total time:

\(2+4=6\text{ h}\).

Average speed:

\(v_{\text{avg}}=\dfrac{360}{6}=60\text{ km/h}\).

Let the distance from the starting point to the waterfalls be \(x\) km.

Then the travel time there is:

\(\dfrac{x}{5}\) h,

and back:

\(\dfrac{x}{4}\) h.

According to the condition, the return trip took \(0.5\) hour longer:

\(\dfrac{x}{4}-\dfrac{x}{5}=0.5\).

\(\dfrac{5x-4x}{20}=0.5\).

\(\dfrac{x}{20}=0.5\).

\(x=10\).

So the tourists walked 10 km one way, and in total:

\(10+10=20\text{ km}\).

After a 20% increase, the price became:

\(200000\cdot1.2=240000\) soums.

Then the price was reduced by 10% of the new price:

\(240000\cdot0.9=216000\) soums.

Therefore, the scarf cost 216,000 soums after the discount.

First find the number of students studying IT:

\(500\cdot0.4=200\).

Now find 60% of 200:

\(200\cdot0.6=120\).

Therefore, 120 students are studying to become programmers.

Find the number of children:

\(60\cdot0.4=24\).

Then the number of adults was:

\(60-24=36\).

A child ticket costs 50% of the adult ticket:

\(30000\cdot0.5=15000\) soums.

Total cost:

for adults: \(36\cdot30000=1080000\) soums,

for children: \(24\cdot15000=360000\) soums.

In total:

\(1080000+360000=1440000\) soums.

First find the number of craftsmen who weave large carpets:

\(200\cdot0.4=80\).

Now find 75% of 80:

\(80\cdot0.75=60\).

Therefore, 60 craftsmen weave large silk carpets.

Let the mass of strawberries be \(x\) g.

Then the mass of bananas is \(2x\) g, since there are twice fewer strawberries.

According to the condition:

\(x+2x=450\).

\(3x=450\).

\(x=150\).

So there were 150 g of strawberries and 300 g of bananas.

To make them equal, the mass of strawberries must become 300 g.

Therefore, we need to add:

\(300-150=150\) g of strawberries.

Let the initial amount of juice be \(x\) ml.

Then the amount of water was \(4x\) ml.

The total was 1 liter, that is 1000 ml:

\(x+4x=1000\).

\(5x=1000\).

\(x=200\).

So initially there were 200 ml of juice and 800 ml of water.

After adding juice, the amount of water stays 800 ml.

The juice should become two times less than the water, i.e. 400 ml.

Therefore, he needs to add:

\(400-200=200\) ml of juice.

Let the second specialist set up \(x\) cash registers per day.

Then the first specialist sets up \(x+2\) cash registers per day.

Their total daily rate is:

\(x+(x+2)=2x+2\).

In 5 days they set up 90 cash registers, so

\(5(2x+2)=90\).

\(2x+2=18\).

\(2x=16\).

\(x=8\).

Therefore, the second specialist sets up 8 cash registers per day, and the first specialist 10 per day.

First find the distance to the next station:

\(s=60\cdot4=240\text{ km}\).

Since the train left 1 hour late but arrived on time, it spent not 4 hours on the trip, but 3 hours.

So its actual average speed was:

\(v=\dfrac{240}{3}=80\text{ km/h}\).