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${\displaystyle x}$ be the number of kilograms he sold in the morning.Then in the afternoon he sold 2x${\displaystyle 2x}$ kilograms. So, the total is x+2x=3x${\displaystyle x+2x=3x}$. This must be equal to 360.
3x=360${\displaystyle 3x=360}$
x=3603${\displaystyle x=\frac{360}{3}}$
x=120${\displaystyle x=120}$
Therefore, the salesman sold 120 kg in the morning and 2⋅120=240${\displaystyle 2\cdot 120=240}$ kg in the afternoon.

Problem 2 Mary, Peter, and Lucy were picking chestnuts. Mary picked twice as much chestnuts than Peter. Boris 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${\displaystyle x}$ be the amount Peter picked. Then Mary and Lucy picked 2x${\displaystyle 2x}$ and x+2${\displaystyle x+2}$, respectively. So
x+2x+x+2=26${\displaystyle x+2x+x+2=26}$
4x=24${\displaystyle 4x=24}$
x=6${\displaystyle x=6}$
Therefore, Peter, Mary, and Lucy picked 6, 12, and 8 kg, respectively.

Problem 3
Sophia finished 23${\displaystyle \frac{2}{3}}$ 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${\displaystyle x}$ be the total number of pages in the book, then she finished 23⋅x${\displaystyle \frac{2}{3}\cdot x}$ pages.
Then she has x−23⋅x=13⋅x${\displaystyle x-\frac{2}{3}\cdot x=\frac{1}{3}\cdot x}$ pages left.
23⋅x−13⋅x=90${\displaystyle \frac{2}{3}\cdot x-\frac{1}{3}\cdot x=90}$
13⋅x=90${\displaystyle \frac{1}{3}\cdot x=90}$
x=270${\displaystyle x=270}$
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${\displaystyle 6}$ tractors ploughed 120${\displaystyle 120}$ hectares a day and they finished the work in 4${\displaystyle 4}$ days, then the whole field is: 120⋅6⋅4=720⋅4=2880${\displaystyle 120\cdot 6\cdot 4=720\cdot 4=2880}$ hectares. Let's suppose that each of the four tractors ploughed x${\displaystyle x}$ hectares a day. Therefore in 5 days they ploughed
5⋅4⋅x=20⋅x${\displaystyle 5\cdot 4\cdot x=20\cdot x}$ hectares, which equals the area of the whole field, 2880 hectares.
So, we get 20x=2880${\displaystyle 20x=2880}$
x=288020=144${\displaystyle x=\frac{2880}{20}=144}$. 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${\displaystyle x}$ be the number he chose, then
2⋅x−138=102${\displaystyle 2\cdot x-138=102}$
2x=240${\displaystyle 2x=240}$
x=120${\displaystyle x=120}$

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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${\displaystyle x}$ be the number I chose, then
x5−154=6${\displaystyle \frac{x}{5}-154=6}$
x5=160${\displaystyle \frac{x}{5}=160}$
x=800${\displaystyle x=800}$

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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⋅t${\displaystyle S=V\cdot t}$

	V (km/hr)	t (hr)	S (km)
Car	x + 5	4	4(x +5)
Truck	X	4	4x

4(x+5)+4x=380${\displaystyle 4(x+5)+4x=380}$
4x+4x=380−20${\displaystyle 4x+4x=380-20}$
8x=360${\displaystyle 8x=360}$
x=3608${\displaystyle x=\frac{360}{8}}$
x=45${\displaystyle x=45}$
Therefore the truck's speed is 45${\displaystyle 45}$ km/hr, and the car's speed is 50${\displaystyle 50}$ 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${\displaystyle x}$ be the length of the longer side x>3${\displaystyle x>3}$, then the other side's length is x−3${\displaystyle x-3}$ cm. Then the area is S₁ = x(x - 3) cm². After we increase the lengths of the sides they will become (x+1)${\displaystyle (x+1)}$ and (x−3+1)=(x−2)${\displaystyle (x-3+1)=(x-2)}$ cm long. Hence the area of the new rectangle will be A2=(x+1)⋅(x−2)${\displaystyle A_2=(x+1)\cdot (x-2)}$ cm², which is 18 cm² more than the first area. Therefore
A1+18=A2${\displaystyle A_1+18=A_2}$
x(x−3)+18=(x+1)(x−2)${\displaystyle x(x-3)+18=(x+1)(x-2)}$
x2−3x+18=x2+x−2x−2${\displaystyle x^2-3x+18=x^2+x-2x-2}$
2x=20${\displaystyle 2x=20}$
x=10${\displaystyle x=10}$. So, the sides of the rectangle are 10${\displaystyle 10}$ cm and 

(10−3)=7${\displaystyle (10-3)=7}$ 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)${\displaystyle (8100-x)}$ 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%${\displaystyle 15\mathrm{\%}}$ or 10%${\displaystyle 10\mathrm{\%}}$.
So 8100+15100⋅x+10100⋅(8100−x)=9100${\displaystyle 8100+\frac{15}{100}\cdot x+\frac{10}{100}\cdot (8100-x)=9100}$
Therefore 8100+320x+110(8100−x)=9100${\displaystyle 8100+\frac{3}{20}x+\frac{1}{10}(8100-x)=9100}$
120x=190${\displaystyle \frac{1}{20}x=190}$
x=3800${\displaystyle x=3800}$
Therefore, the cows produced 3800 and 4300 litres of milk the first year, and 4370${\displaystyle 4370}$ and 4730${\displaystyle 4730}$ 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)${\displaystyle (148-x)}$ km. By the time of the meeting at station C, the express train travelled for 148−x80+1060${\displaystyle \frac{148-x}{80}+\frac{10}{60}}$ hours and the freight train travelled for x36+560${\displaystyle \frac{x}{36}+\frac{5}{60}}$ hours. The trains left at the same time, so: 148−x80+16=x36+112${\displaystyle \frac{148-x}{80}+\frac{1}{6}=\frac{x}{36}+\frac{1}{12}}$. The common denominator for 6, 12, 36, 80 is 720. Then
9(148−x)+120=20x+60${\displaystyle 9(148-x)+120=20x+60}$
1332−9x+120=20x+60${\displaystyle 1332-9x+120=20x+60}$
29x=1392${\displaystyle 29x=1392}$
x=48${\displaystyle x=48}$. 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 4836+560${\displaystyle \frac{48}{36}+\frac{5}{60}}$ hours, i.e. 1${\displaystyle 1}$ hour and 25${\displaystyle 25}$ min.
Therefore it left station B at 12−(1+2560)=10+3560${\displaystyle 12-(1+\frac{25}{60})=10+\frac{35}{60}}$ 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${\displaystyle x}$ be the distance between A and B. Since Susan covered 80 km in 2 hours, her speed was V=802=40${\displaystyle V=\frac{80}{2}=40}$km/hr.
If she continued at the same speed she would be 15${\displaystyle 15}$ minutes late, i.e. the planned time on the road is x40−1560${\displaystyle \frac{x}{40}-\frac{15}{60}}$hr. The rest of the distance is (x−80)${\displaystyle (x-80)}$ km. V=40+10=50${\displaystyle V=40+10=50}$ km/hr.
So, she covered the distance between A and B in 2+x−8050${\displaystyle 2+\frac{x-80}{50}}$ hr, and it was 36 min less than planned. Therefore, the planned time was 2+x−8050+3660${\displaystyle 2+\frac{x-80}{50}+\frac{36}{60}}$.
When we equalize the expressions for the scheduled time, we get the equation:
x40−1560=2+x−8050+3660${\displaystyle \frac{x}{40}-\frac{15}{60}=2+\frac{x-80}{50}+\frac{36}{60}}$
x−1040=100+x−80+3050${\displaystyle \frac{x-10}{40}=\frac{100+x-80+30}{50}}$
x−104=x+505${\displaystyle \frac{x-10}{4}=\frac{x+50}{5}}$
5x−50=4x+200${\displaystyle 5x-50=4x+200}$
x=250${\displaystyle x=250}$
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${\displaystyle x}$ 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⋅25+(x−3)⋅30=75+30(x−3)${\displaystyle 3\cdot 25+(x-3)\cdot 30=75+30(x-3)}$
Therefore: 25x=75+30(x−3)−100${\displaystyle 25x=75+30(x-3)-100}$
25x=75+30x−90−100${\displaystyle 25x=75+30x-90-100}$
190−75=30x−25${\displaystyle 190-75=30x-25}$
115=5x${\displaystyle 115=5x}$
x=23${\displaystyle x=23}$
So the company worked 23 days and they made 23⋅25+100=675${\displaystyle 23\cdot 25+100=675}$ 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${\displaystyle 24}$ plants. How many birches and roses were planted?

Solution:
Let x${\displaystyle x}$ be the number of roses. Then the number of birches is 24−x${\displaystyle 24-x}$, and the number of boys is 3×(24−x)${\displaystyle 3\times (24-x)}$. If each girl planted 3 roses, there are x3${\displaystyle \frac{x}{3}}$ girls in the class.
We know that there are 24 students in the class. Therefore x3+3(24−x)=24${\displaystyle \frac{x}{3}+3(24-x)=24}$
x+9(24−x)=3⋅24${\displaystyle x+9(24-x)=3\cdot 24}$
x+216−9x=72${\displaystyle x+216-9x=72}$
216−72=8x${\displaystyle 216-72=8x}$
1448=x${\displaystyle \frac{144}{8}=x}$
x=18${\displaystyle x=18}$
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.
1st case. The stop was planned. Let us consider only the trip from C to B, and let x${\displaystyle x}$ be the number of hours the driver spent on this trip.
Then the distance from C to B is S=40⋅x${\displaystyle S=40\cdot x}$ km. If the driver could use the initial route, it would take him x−3060=x−12${\displaystyle x-\frac{30}{60}=x-\frac{1}{2}}$ hours to drive from C to B. The distance from C to B according to the initially itinerary was (x−12)⋅32${\displaystyle (x-\frac{1}{2})\cdot 32}$ km, and this distance is 28${\displaystyle 28}$ km shorter than 40⋅x${\displaystyle 40\cdot x}$ km. Then we have the equation
(x−1/2)⋅32+28=40x${\displaystyle (x-1/2)\cdot 32+28=40x}$
32x−16+28=40x${\displaystyle 32x-16+28=40x}$
8x=12${\displaystyle 8x=12}$
x=128${\displaystyle x=\frac{12}{8}}$   x=1412=12060=${\displaystyle x=1\frac{4}{12}=1\frac{20}{60}=}$ 1 hr 20 min.
So, the car covered the distance between C and B in 1 hour and 20 min.
The distance from A to B is 3⋅32+128⋅40=96+60=156${\displaystyle 3\cdot 32+\frac{12}{8}\cdot 40=96+60=156}$ km.
2nd case. The driver did not plan the stop at C. Suppose it took x${\displaystyle x}$ hours for him to get from C to B. Then the distance is S=40⋅x${\displaystyle S=40\cdot x}$ km. It took x−3060−1560=x−4560=x−34${\displaystyle x-\frac{30}{60}-\frac{15}{60}=x-\frac{45}{60}=x-\frac{3}{4}}$ h to drive from C to B. The distance from C to B is 32(x−34)${\displaystyle 32(x-\frac{3}{4})}$ km, which is 28${\displaystyle 28}$ km shorter than 40⋅x${\displaystyle 40\cdot x}$, i.e.
32(x−34)+28=40x${\displaystyle 32(x-\frac{3}{4})+28=40x}$
32x−24+28=40x${\displaystyle 32x-24+28=40x}$
4=8x${\displaystyle 4=8x}$
x=12hr⋅x=30min.${\displaystyle x=\frac{1}{2}\text{hr}\cdot x=30\text{min}.}$ Then the time of the trip from C to B was 30 min. The distance covered equals 3⋅32+12⋅40=96+20=116km${\displaystyle 3\cdot 32+\frac{1}{2}\cdot 40=96+20=116km}$.

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${\displaystyle x}$ be the number of days in the initial plan. Therefore, the whole field is 120⋅x${\displaystyle 120\cdot x}$ hectares. The farmer had to work for x+2${\displaystyle x+2}$ days, and he ploughed 85(x+2)${\displaystyle 85(x+2)}$ hectares, leaving 40${\displaystyle 40}$ hectares unploughed. Then we have the equation:
120x=85(x+2)+40${\displaystyle 120x=85(x+2)+40}$
35x=210${\displaystyle 35x=210}$
x=6${\displaystyle x=6}$
So the farmer planned to have the work done in 6 days, and the area of the farm field is 120⋅6=720${\displaystyle 120\cdot 6=720}$ hectares.