Math Word Problems and Solutions - Distance, Speed, Time
Problem 1
A salesman sold twice as much pears in the afternoon than in the morning.
If he sold 360 kilograms of pears that day, how many
kilograms did he sell in the morning and how many in the afternoon?
Solution: Let $x$ be the number of kilograms he
sold in the morning.Then in the afternoon he sold $2x$ kilograms. So, the
total is $x + 2x = 3x$. This must be equal to 360. $3x = 360$ $x = \frac{360}{3}$ $x = 120$
Therefore, the salesman sold 120 kg in the morning and $2\cdot 120 = 240$ kg in the afternoon.
Problem 2 Mary, Peter, and Lucy were picking chestnuts. Mary picked twice as much chestnuts than Peter. Lucy picked
2 kg more than Peter. Together the three of them picked 26 kg of chestnuts. How many kilograms did each of them pick?
Solution: Let $x$ be the amount Peter picked. Then Mary and Lucy picked $2x$ and $x+2$, respectively.
So $x+2x+x+2=26$ $4x=24$ $x=6$ Therefore, Peter, Mary, and Lucy picked 6, 12, and 8 kg, respectively.
Problem 3 Sophia finished $\frac{2}{3}$ of a book. She calculated that she finished 90 more pages than she has yet to read. How long is her book?
Solution: Let $x$ be the total number of pages in the book, then she finished $\frac{2}{3}\cdot x$ pages.
Then she has $x-\frac{2}{3}\cdot x=\frac{1}{3}\cdot x$ pages left. $\frac{2}{3}\cdot x-\frac{1}{3}\cdot x=90$
$\frac{1}{3}\cdot x=90$
$x=270$ So the book is 270 pages long.
Problem 4 A farming field can be ploughed by 6 tractors in 4 days. When 6 tractors work together, each of them ploughs
120 hectares a day. If two of the tractors were moved to another field,
then the remaining 4 tractors could plough the same field in 5 days.
How many hectares a day would one tractor plough then?
Solution:
If each of $6$ tractors ploughed $120$ hectares a day and they finished the work in $4$
days, then the whole field is: $120\cdot 6 \cdot 4 = 720 \cdot 4 = 2880$ hectares. Let's
suppose that each of the four tractors ploughed $x$ hectares a day. Therefore in 5 days they ploughed
$5 \cdot 4 \cdot x = 20 \cdot x$ hectares, which equals the area of the whole field, 2880 hectares.
So, we get $20x = 2880$
$ x = \frac{2880}{20} = 144$. Hence, each of the four tractors would plough 144 hectares a day.
Problem 5
A student chose a number, multiplied it by 2, then subtracted 138 from the result and got 102. What was the number he chose?
Solution: Let $x$ be the number he chose, then
$2\cdot x - 138 = 102$ $2x = 240$ $x = 120$
Problem 6
I chose a number and divide it by 5. Then I subtracted 154 from the result and got 6. What was the number I chose?
Solution: Let $x$ be the number I chose, then
$\frac{x}{5}-154=6$ $\frac{x}{5}=160$ $x=800$
Problem 7
The distance between two towns is 380 km. At the same moment, a passenger car and a truck start moving towards each other from
different towns. They meet 4 hours later. If the car drives 5 km/hr faster than the truck, what are their speeds?
Solution:
The main idea used in this kind of problems is that the distance equals speed multiplied by time $S = V\cdot t$
V (km/hr)
t (hr)
S (km)
Car
x + 5
4
4(x +5)
Truck
X
4
4x
$4(x + 5) + 4x = 380$ $4x + 4x = 380 - 20$ $8x = 360$ $x = \frac{360}{8}$ $x = 45$
Therefore the truck's speed is $45$ km/hr, and the car's speed is $50$ km/hr.
Problem 8
One side of a rectangle is 3 cm shorter than the other side. If we increase the length of each side by 1 cm, then the area of the rectangle
will increase by 18 cm2. Find the lengths of all sides.
Solution: Let $x$ be the length of the longer side $x \gt 3$, then the other side's length is $x-3$ cm. Then the area is S1 = x(x - 3) cm2.
After we increase the lengths of the sides they will become $(x +1)$ and $(x - 3 + 1) = (x - 2)$ cm long. Hence the area of the
new rectangle will be $A_2 = (x + 1)\cdot(x - 2)$ cm2, which is 18 cm2 more than the first area. Therefore
$A_1 +18 = A_2$ $x(x - 3) + 18 = (x + 1)(x - 2)$ $x^2 - 3x + 18 = x^2 + x - 2x - 2$ $2x = 20$ $x = 10$.
So, the sides of the rectangle are $10$ cm and $(10 - 3) = 7$ cm long.
Problem 9 The first year, two cows produced 8100 litres of milk. The second year their production increased
by 15% and 10% respectively, and the total amount of milk increased to
9100 litres a year. How many litres were milked from each cow each year?
Solution: Let x be the amount of milk the first cow
produced during the first year. Then the second cow produced $(8100 - x)$ litres of milk that year. The second year, each cow produced the
same amount of milk as they did the first year plus the increase of $15\%$ or $10\%$.
So $8100 + \frac{15}{100}\cdot x + \frac{10}{100} \cdot (8100 - x) = 9100$
Therefore $8100 + \frac{3}{20}x + \frac{1}{10}(8100 - x) = 9100$ $\frac{1}{20}x = 190$ $x = 3800$ Therefore,
the cows produced 3800 and 4300 litres of milk the first year, and $4370$ and $4730$ litres of milk the second year, respectively.
Problem 10 The
distance between stations A and B is 148 km. An express train left station A towards station B with the speed of 80 km/hr. At the same
time, a freight train left station B towards station A with the speed of 36 km/hr. They met at station C at 12 pm, and by that time the
express train stopped at at intermediate station for 10 min and the freight train stopped for 5 min. Find:
a) The distance between stations C and B.
b) The time when the freight train left station B.
Solution a) Let x be the distance between
stations B and C. Then the distance from station C to station A is $(148 - x)$ km. By the time of the meeting at station C, the express train
travelled for $\frac{148-x}{80}+\frac{10}{60}$ hours and the freight train travelled for $\frac{x}{36}+\frac{5}{60}$ hours. The trains left at the same time, so:
$\frac{148 - x}{80} + \frac{1}{6} = \frac{x}{36} + \frac{1}{12}$. The common denominator for 6, 12, 36, 80 is 720. Then
$9(148 - x) +120 = 20x +60$ $1332 - 9x + 120 = 20x + 60$ $29x = 1392$ $x = 48$.
Therefore the distance between stations B and C is 48 km.
b) By the time of the meeting at station C the freight
train rode for $\frac{48}{36} + \frac{5}{60}$ hours, i.e. $1$ hour and $25$ min. Therefore it left station B at $12 - (1 + \frac{25}{60}) = 10 + \frac{35}{60}$ hours, i.e. at 10:35 am.
Problem 11 Susan drives from city A to city B. After two hours of driving she
noticed that she covered 80 km and calculated that, if she continued
driving at the same speed, she would end up been 15 minutes late. So
she increased her speed by 10 km/hr and she arrived at city B 36 minutes earlier
than she planned.
Find the distance between cities A and B.
Solution:
Let $x$ be the distance between A and B. Since Susan covered 80 km in 2 hours, her speed was $V = \frac{80}{2} = 40$ km/hr.
If she continued at the same speed she would be $15$ minutes late, i.e. the planned time on the road is $\frac{x}{40} - \frac{15}{60}$ hr.
The rest of the distance is $(x - 80)$ km. $V = 40 + 10 = 50$ km/hr.
So, she covered the distance between A and B in $2 +\frac{x - 80}{50}$ hr, and it was 36 min less than planned.
Therefore, the planned time was $2 + \frac{x -80}{50} + \frac{36}{60}$.
When we equalize the expressions for the scheduled time, we get the equation:
$\frac{x}{40} - \frac{15}{60} = 2 + \frac{x -80}{50} + \frac{36}{60}$
$\frac{x - 10}{40} = \frac{100 + x - 80 + 30}{50}$
$\frac{x - 10}{4} = \frac{x +50}{5}$
$5x - 50 = 4x + 200$
$x = 250$
So, the distance between cities A and B is 250 km.
Problem 12 To deliver an order on time, a company has to make 25 parts a day. After making 25 parts per day for 3
days, the company started to produce 5 more parts per day, and by the last day of work 100 more parts than planned were produced.
Find how many parts the company made and how many days this took.
Solution:
Let $x$ be the number of days the company worked. Then 25x is the
number of parts they planned to make. At the new production rate they
made:
$3\cdot 25 + (x - 3)\cdot 30 = 75 + 30(x - 3)$
Therefore: $25 x = 75 + 30(x -3) - 100$
$25x = 75 +30x -90 - 100$
$190 -75 = 30x -25$
$115 = 5x$
$x = 23$
So the company worked 23 days and they made $23\cdot 25+100 = 675$ pieces.
Problem 13
There are 24 students in a seventh grade class. They decided to plant birches and roses at the school's backyard. While each girl planted 3
roses, every three boys planted 1 birch. By the end of the day they planted $24$ plants. How many birches and roses were planted?
Solution:
Let $x$ be the number of roses. Then the number of birches is $24 - x$, and the number of boys is $3\times (24-x)$. If each girl planted 3
roses, there are $\frac{x}{3}$ girls in the class. We know that there are 24 students in the class. Therefore $\frac{x}{3} + 3(24 - x) = 24$
$x + 9(24 - x) = 3\cdot 24$
$x +216 - 9x = 72$ $216 - 72 = 8x$ $\frac{144}{8} = x$ $x = 18$
So, students planted 18 roses and 24 - x = 24 - 18 = 6 birches.
Problem 14
A car left town A towards town B driving at a speed of V = 32 km/hr. After 3 hours on the road the driver stopped for 15 min in town C. Because of a
closed road he had to change his route, making the trip 28 km longer. He increased his speed to V = 40 km/hr but still he was 30 min late. Find:
a) The distance the car has covered.
b) The time that took it to get from C to B.
Solution: From the statement of the problem we don't know if the 15 min stop in town C was planned or it was
unexpected. So we have to consider both cases.
A The stop was planned. Let us consider only the trip from C to B, and let $x$ be the number of hours the driver
spent on this trip.
Then the distance from C to B is $S = 40\cdot x$
km. If the driver could use the initial route, it would take him $x - \frac{30}{60} = x - \frac{1}{2}$ hours to drive from C to B. The distance from C to B
according to the initially itinerary was $(x - \frac{1}{2})\cdot 32$ km, and this
distance is $28$ km shorter than $40\cdot x$ km. Then we have the equation
$(x - 1/2)\cdot 32 + 28 = 40x$
$32x -16 +28 = 40x$
$-8x = -12$ $8x = 12$
$x = \frac{12}{8}$
$x = 1 \frac{4}{8} = 1 \frac{1}{2} = 1 \frac{30}{60} =$ 1 hr 30 min.
So, the car covered the distance between C and B in 1 hour and 30 min. The distance from A to B is $3\cdot 32 + \frac{12}{8}\cdot 40 = 96 + 60 = 156$ km.
B Suppose it took $x$ hours for him
to get from C to B. Then the distance is $S = 40\cdot x$ km.
The driver did not plan the stop at C. Let we accept that he stopped because he had to change the route.
It took $x - \frac{30}{60} + \frac{15}{60} = x - \frac{15}{60} = x - \frac{1}{4}$ h to drive from C to B. The
distance from C to B is $32(x - \frac{1}{4})$ km, which is $28$ km shorter than $40\cdot x$, i.e.
$32(x - \frac{1}{4}) + 28 = 40x$ $32x - 8 +28 = 40x$ $20= 8x$
$x = \frac{20}{8} = \frac{5}{2} = 2 \text{hr } 30 \text{min}.$
The distance covered equals $ 40 \times 2.5 = 100 km$.
Problem 15
If a farmer wants to plough a farm field on time, he must plough 120 hectares a day. For technical reasons he ploughed only 85 hectares a day, hence he had to plough 2 more days than he planned and he
still has 40 hectares left. What is the area of the farm field and how many days the farmer planned to work initially?
Solution:
Let $x$ be the number of days in the initial plan. Therefore, the whole field is $120\cdot x$ hectares. The farmer had to work for $x + 2$ days, and he
ploughed $85(x + 2)$ hectares, leaving $40$ hectares unploughed. Then we have the equation:
$120x = 85(x + 2) + 40$
$35x = 210$
$x = 6$
So the farmer planned to have the work done in 6 days, and the area of the farm field is $120\cdot 6 = 720$ hectares.
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 $x$ be the number of parts the woodworker normally makes daily. In 24 days he makes $24\cdot x$ pieces. His new daily production rate is $x + 5$ pieces and in
$22$ days he made $22 \cdot (x + 5)$ parts. This is 80 more than $24\cdot x$. Therefore
the equation is: $24\cdot x + 80 = 22(x +5)$ $30 = 2x$ $x = 15$
Normally he makes 15 parts a day and in 24 days he makes $15 \cdot 24 = 360$ 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 $2\frac{30}{60} \cdot x$ km and $2\frac{20}{60} \cdot (x + 2)$ km. From the equation: $2\frac{30}{60} \cdot x = 2\frac{20}{60} \cdot (x+2)$
we get $x = 28$ km/hr.
The intial speed of the biker is 28 km/h.
Half the distance between the two towns is
$2 h 30 min \times 28 = 2.5 \times 28 = 70$.
So the distance is $2 \times 70 = 140$ 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 $\frac{5}{3}$ 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 $\frac{5}{3}$ m/sec $= \frac{5\cdot 60\cdot 60}{\frac{3}{1000}}$ km/hr = $6$ km/hr. Therefore, the
new speed is $48 + 6 = 54$ km/hr. If it took $x$ hours to cover the first
half of the distance, then it took $x - \frac{15}{60} = x - 0.25$ hr to cover the
second part. So the equation is: $48 \cdot x = 54 \cdot (x - 0.25)$
$48 \cdot x = 54 \cdot x - 54\cdot 0.25$
$48 \cdot x - 54 \cdot x = - 13.5$
$-6x = - 13.5$
$x = 2.25$ h.
The whole distance is
$2 \times 48 \times 2.25 = 216$ 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 $\frac{1}{15}$ of the job per day and Tony does $75\%$ of $\frac{1}{15}$, i.e.
$\frac{75}{100}\cdot \frac{1}{15} = \frac{1}{20}$. Suppose that Tony worked alone
for $x$ days. Then he finished $\frac{x}{20}$ of the total job alone. Working
together for 6 days, the two workers finished $6\cdot (\frac{1}{15}+\frac{1}{20}) = 6\cdot \frac{7}{60} = \frac{7}{10}$ of the job.
The sum of $\frac{x}{20}$ and $\frac{7}{10}$ gives us the whole job, i.e. $1$. So we get the equation:
$\frac{x}{20}+\frac{7}{10}=1$
$\frac{x}{20} = \frac{3}{10}$
$x = 6$. Tony worked for 6 + 6 = 12 days
and Elizabeth worked for $6$ days. The part of job done
is $12\cdot \frac{1}{20} = \frac{60}{100} = 60\%$ for Tony, and $6\cdot \frac{1}{15} = \frac{40}{100} = 40\%$ 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
$\frac{25}{100} \cdot 120 = 30$ hectares, therefore $120 + 30 = 150$ hectares is the
new daily productivity. Lets x be the planned number of
days allotted for the job. Then the farm is $120\cdot x$ hectares. On the
other hand, we get the same area if we add $120 \cdot 2$ hectares to
$150(x -4)$ hectares. Then we get the equation
$120x = 120\cdot 2 + 150(x -4)$
$x = 12$
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 $120 \cdot 12 = 1440$ 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
$33 \times \frac{1}{3}\%$, 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 $x$ 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 $75(x - \frac{48}{60})$ km. On the other hand, it is $50x + 40$ km. So we get the equation:
$75(x - \frac{48}{60}) = 50x + 40$ $x = 4$ hr is the scheduled travel time. The
distance between the two stations is $50\cdot 4 +40 = 240$ km. Then the speed the train must keep to be on schedule is $\frac{240}{4} = 60$ 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 $x$ km/hr. Then the speed of
the faster train is $(x + 10)$ km/hr. In 2 hours they cover $2x$ km and $2(x +10)$km, respectively. Therefore if they didn't meet yet, the whole
distance from A to B is $2x + 2(x +10) +40 = 4x +60$ km. However, if
they already met and continued to move, the distance would be $2x + 2(x + 10) - 40 = 4x - 20$km.
So we get the following equations: $4x + 60 = 300$ $4x = 240$ $x = 60$ or
$4x - 20 = 300$ $4x = 320$ $x = 80$ Hence the speed of the slower train is $60$ km/hr or $80$ km/hr and the speed of
the faster train is $70$ km/hr or $90$ 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 $\frac{50}{9}$ 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 $\frac{50}{9}$ m/sec $= \frac{50\cdot60\cdot60}{\frac{9}{1000}}$ km/hr $= 20$ km/hr. Therefore, the new speed is $V = 50 + 20 = 70$ km/hr. If $x$ 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 $(x +\frac{42}{60})$ hr. When the speed of the bus is $V = 70$ km/hr, the travel time is $x - \frac{30}{60}$ hr. Then
$50(x +\frac{42}{60}) = 70(x-\frac{30}{60})$ $5(x+\frac{7}{10}) = 7(x-\frac{1}{2})$
$\frac{7}{2} + \frac{7}{2} = 7x -5x$
$2x = 7$ $x = \frac{7}{2}$ hr. So, the bus is scheduled to make the trip in $3$ hr $30$ min.
The distance between the two towns is $70(\frac{7}{2} - \frac{1}{2}) = 70\cdot 3 = 210$ km and the scheduled speed is $\frac{210}{\frac{7}{2}} = 60$ km/hr.
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