Math Word Problems and Solutions - Distance, Speed, Time

Problem 16
A woodworker normally makes a certain number of parts in 24 days. But he was able to increase his productivity by 5 parts per day, and so he not only finished the job in only 22 days but also he made 80 extra parts. How many parts does the woodworker normally makes per day and how many pieces does he make in 24 days?

Solution:
Let x${\displaystyle x}$ be the number of parts the woodworker normally makes daily. In 24 days he makes 24⋅x${\displaystyle 24\cdot x}$ pieces. His new daily production rate is x+5${\displaystyle x+5}$ pieces and in 22${\displaystyle 22}$ days he made 22⋅(x+5)${\displaystyle 22\cdot (x+5)}$ parts. This is 80 more than 24⋅x${\displaystyle 24\cdot x}$. Therefore the equation is:
24⋅x+80=22(x+5)${\displaystyle 24\cdot x+80=22(x+5)}$
30=2x${\displaystyle 30=2x}$
x=15${\displaystyle x=15}$
Normally he makes 15 parts a day and in 24 days he makes 15⋅24=360${\displaystyle 15\cdot 24=360}$ parts.

Problem 17
A biker covered half the distance between two towns in 2 hr 30 min. After that he increased his speed by 2 km/hr. He covered the second half of the distance in 2 hr 20 min. Find the distance between the two towns and the initial speed of the biker.

Solution:

Let x km/hr be the initial speed of the biker, then his speed during the second part of the trip is x + 2 km/hr. Half the distance between two cities equals 23060⋅x${\displaystyle 2\frac{30}{60}\cdot x}$ km and 22060⋅(x+2)${\displaystyle 2\frac{20}{60}\cdot (x+2)}$ km. From the equation: 23060⋅x=22060⋅(x+2)${\displaystyle 2\frac{30}{60}\cdot x=2\frac{20}{60}\cdot (x+2)}$ we get x=28${\displaystyle x=28}$ km/hr.
Therefore the distance between the two towns is equal to 2⋅2⋅2060⋅28=140${\displaystyle \frac{2\cdot 2\cdot 20}{60\cdot 28}=140}$ km.

Problem 18
A train covered half of the distance between stations A and B at the speed of 48 km/hr, but then it had to stop for 15 min. To make up for the delay, it increased its speed by 53${\displaystyle \frac{5}{3}}$ m/sec and it arrived to station B on time. Find the distance between the two stations and the speed of the train after the stop.

Solution:
First let us determine the speed of the train after the stop. The speed was increased by 53${\displaystyle \frac{5}{3}}$ m/sec =5⋅60⋅6031000${\displaystyle =\frac{5\cdot 60\cdot 60}{\frac{3}{1000}}}$km/hr = 6${\displaystyle 6}$ km/hr. Therefore, the new speed is 48+6=54${\displaystyle 48+6=54}$ km/hr. If it took x${\displaystyle x}$ hours to cover the first half of the distance, then it took x−1560=x−0.25${\displaystyle x-\frac{15}{60}=x-0.25}$ hr to cover the second part.
So the equation is: 48⋅x=54⋅(x−0.25)${\displaystyle 48\cdot x=54\cdot (x-0.25)}$
x=13.5${\displaystyle x=13.5}$ h. The whole distance is 2×48×13.5=1296${\displaystyle 2\times 48\times 13.5=1296}$ km.

Problem 19
Elizabeth can get a certain job done in 15 days, and Tony can finish only 75% of that job within the same time. Tony worked alone for several days and then Elizabeth joined him, so they finished the rest of the job in 6 days, working together.
For how many days have each of them worked and what percentage of the job have each of them completed?

Solution:
First we will find the daily productivity of every worker. If we consider the whole job as unit (1), Elizabeth does 115${\displaystyle \frac{1}{15}}$ of the job per day and Tony does 75%${\displaystyle 75\mathrm{\%}}$ of 115${\displaystyle \frac{1}{15}}$, i.e.
75100⋅115=120${\displaystyle \frac{75}{100}\cdot \frac{1}{15}=\frac{1}{20}}$. Suppose that Tony worked alone for x${\displaystyle x}$ days. Then he finished x20${\displaystyle \frac{x}{20}}$ of the total job alone. Working together for 6 days, the two workers finished 6⋅(115+120)=6⋅760=710${\displaystyle 6\cdot (\frac{1}{15}+\frac{1}{20})=6\cdot \frac{7}{60}=\frac{7}{10}}$ of the job.
The sum of x20${\displaystyle \frac{x}{20}}$ and 710${\displaystyle \frac{7}{10}}$ gives us the whole job, i.e. 1${\displaystyle 1}$. So we get the equation:
x20+710=1${\displaystyle \frac{x}{20}+\frac{7}{10}=1}$
x20=310${\displaystyle \frac{x}{20}=\frac{3}{10}}$
x=6${\displaystyle x=6}$. Tony worked for 6 + 6 = 12 days and Elizabeth worked for 6${\displaystyle 6}$ days. The part of job done is 12⋅120=60100=60%${\displaystyle 12\cdot \frac{1}{20}=\frac{60}{100}=60\mathrm{\%}}$ for Tony, and 6⋅115=40100=40%${\displaystyle 6\cdot \frac{1}{15}=\frac{40}{100}=40\mathrm{\%}}$ for Elizabeth.

Problem 20
A farmer planned to plough a field by doing 120 hectares a day. After two days of work he increased his daily productivity by 25% and he finished the job two days ahead of schedule.
a) What is the area of the field?
b) In how many days did the farmer get the job done?
c) In how many days did the farmer plan to get the job done?

Solution:
First of all we will find the new daily productivity of the farmer in hectares per day: 25% of 120 hectares is 25100⋅120=30${\displaystyle \frac{25}{100}\cdot 120=30}$ hectares, therefore 120+30=150${\displaystyle 120+30=150}$ hectares is the new daily productivity. Lets x be the planned number of days allotted for the job. Then the farm is 120⋅x${\displaystyle 120\cdot x}$ hectares. On the other hand, we get the same area if we add 120⋅2${\displaystyle 120\cdot 2}$ hectares to 150(x−4)${\displaystyle 150(x-4)}$ hectares. Then we get the equation
120x=120⋅2+150(x−4)${\displaystyle 120x=120\cdot 2+150(x-4)}$
x=12${\displaystyle x=12}$
So, the job was initially supposed to take 12 days, but actually the field was ploughed in 12 - 2 =10 days. The field's area is 120⋅12=1440${\displaystyle 120\cdot 12=1440}$ hectares.

Problem 21
To mow a grass field a team of mowers planned to cover 15 hectares a day. After 4 working days they increased the daily productivity by 33×13%${\displaystyle 33\times \frac{1}{3}\mathrm{\%}}$, and finished the work 1 day earlier than it was planned.
A) What is the area of the grass field?
B) How many days did it take to mow the whole field?
C) How many days were scheduled initially for this job?
Hint: See problem 20 and solve by yourself.
Answer: A) 120 hectares; B) 7 days; C) 8 days.

Problem 22
A train travels from station A to station B. If the train leaves station A and makes 75 km/hr, it arrives at station B 48 minutes ahead of scheduled. If it made 50 km/hr, then by the scheduled time of arrival it would still have 40 km more to go to station B. Find:
A) The distance between the two stations;
B) The time it takes the train to travel from A to B according to the schedule;
C) The speed of the train when it's on schedule.

Solution:
Let x${\displaystyle x}$ be the scheduled time for the trip from A to B. Then the distance between A and B can be found in two ways. On one hand, this distance equals 75(x−4860)${\displaystyle 75(x-\frac{48}{60})}$ km. On the other hand, it is 50x+40${\displaystyle 50x+40}$ km. So we get the equation:
75(x−4860)=50x+40${\displaystyle 75(x-\frac{48}{60})=50x+40}$
x=4${\displaystyle x=4}$ hr is the scheduled travel time. The distance between the two stations is 50⋅4+40=240${\displaystyle 50\cdot 4+40=240}$ km. Then the speed the train must keep to be on schedule is 2404=60${\displaystyle \frac{240}{4}=60}$ km/hr.

Problem 23
The distance between towns A and B is 300 km. One train departs from town A and another train departs from town B, both leaving at the same moment of time and heading towards each other. We know that one of them is 10 km/hr faster than the other. Find the speeds of both trains if 2 hours after their departure the distance between them is 40 km.

Solution:
Let the speed of the slower train be x${\displaystyle x}$ km/hr. Then the speed of the faster train is (x+10)${\displaystyle (x+10)}$ km/hr. In 2 hours they cover 2x${\displaystyle 2x}$ km and 2(x+10)${\displaystyle 2(x+10)}$km, respectively. Therefore if they didn't meet yet, the whole distance from A to B is 2x+2(x+10)+40=4x+60${\displaystyle 2x+2(x+10)+40=4x+60}$ km. However, if they already met and continued to move, the distance would be 2x+2(x+10)−40=4x−20${\displaystyle 2x+2(x+10)-40=4x-20}$km. So we get the following equations:
4x+60=300${\displaystyle 4x+60=300}$
4x=240${\displaystyle 4x=240}$
x=60${\displaystyle x=60}$ or
4x−20=300${\displaystyle 4x-20=300}$
4x=320${\displaystyle 4x=320}$
x=80${\displaystyle x=80}$
Hence the speed of the slower train is 60${\displaystyle 60}$ km/hr or 80${\displaystyle 80}$ km/hr and the speed of the faster train is 70${\displaystyle 70}$ km/hr or 90${\displaystyle 90}$km/hr.

Problem 24
A bus travels from town A to town B. If the bus's speed is 50 km/hr, it will arrive in town B 42 min later than scheduled. If the bus increases its speed by 509${\displaystyle \frac{50}{9}}$ m/sec, it will arrive in town B 30 min earlier than scheduled. Find:
A) The distance between the two towns;
B) The bus's scheduled time of arrival in B;
C) The speed of the bus when it's on schedule.

Solution:
First we will determine the speed of the bus following its increase. The speed is increased by 509${\displaystyle \frac{50}{9}}$ m/sec =50⋅60⋅6091000${\displaystyle =\frac{50\cdot 60\cdot 60}{\frac{9}{1000}}}$ km/hr =20${\displaystyle =20}$ km/hr. Therefore, the new speed is V=50+20=70${\displaystyle V=50+20=70}$ km/hr. If x${\displaystyle x}$ is the number of hours according to the schedule, then at the speed of 50 km/hr the bus travels from A to B within (x+4260)${\displaystyle (x+\frac{42}{60})}$ hr. When the speed of the bus is V=70${\displaystyle V=70}$ km/hr, the travel time is x−3060${\displaystyle x-\frac{30}{60}}$ hr. Then
50(x+4260)=70(x−3060)${\displaystyle 50(x+\frac{42}{60})=70(x-\frac{30}{60})}$
5(x+710)=7(x−12)${\displaystyle 5(x+\frac{7}{10})=7(x-\frac{1}{2})}$
72+72=7x−5x${\displaystyle \frac{7}{2}+\frac{7}{2}=7x-5x}$
2x=7${\displaystyle 2x=7}$
x=72${\displaystyle x=\frac{7}{2}}$ hr.
So, the bus is scheduled to make the trip in 3${\displaystyle 3}$ hr 30${\displaystyle 30}$ min.
The distance between the two towns is 70(72−12)=70⋅3=210${\displaystyle 70(\frac{7}{2}-\frac{1}{2})=70\cdot 3=210}$ km and the scheduled speed is 21072=60${\displaystyle \frac{210}{\frac{7}{2}}=60}$km/hr.