# Maths Word Problems and Solutions

Problem 1 In one store in the afternoon sold twice more pears, than in the morning. During the whole day they sold 360 kg. pears. How many kilograms of pears are sold in the morning and how many in the afternoon?
Solution:
Lets take that the sold in the morning pears are x kg, then in the afternoon are sold 2x kg. Their sum x + 2x = 3x kg is the whole quantity sold pears, 360 kg. So we get the following equation
3x = 360 <=> x = 360/3 <=> x = 120
Therefore in the morning they sold 120 kg pears, and in the afternoon 2.120 = 240 kg.

Problem 2 Ivan gathered twice more chestnuts than Peter and Boris gathered 2kg. more than Peter. Together they gathered 26 kg. chestnuts. How many kilogrammes gathered each one of them?
Solution:
Lets take Peters chestnuts as x kg., then Ivan gathered 2x kg. and Boris (x +2) kg . All gathered chestnuts are: x + 2x +x +2 = 4x +2 and by condition they are 26 kg. We get the equation: 4x +2 = 26 <=> 4x = 24 <=> x = 6
Therefore Peter gathered 6 kg., Ivan 2.6 = 12kg, Boris 6 +2 = 8 kg chestnuts.

Problem 3
Kamen read 2/3 of a book and calculated that the read part is with 90 pages more than the unread. How many pages is the whole book?
Solution:
Lets take the whole book as x pages. The part he read is 2/3 from x , i.e. 2/3.x We will get the unread part when from the whole book we subtract the read part, i.e. x - 2/3 . x = 3/3x - 2/3x = 1/3x The read part 2/3x is with 90 pages more than the unread one, which is 1/3x Therefore
2/3x  1/3x = 90 <=> 1/3x =90 <=> x = 90.3 = 270 So the book is 270 pages.

Problem 4
One tract can be ploughed with 6 tractors for 4 days, if they plough 120hectares a day. Two of the tractors were moved to another tract. The rest 4 ploughed the same tract for 5 days. How many decares average a day ploughed the 4 tractors?
Solution:
If 6 tractors ploughed 120 hectares a day and finished the tract for 4 days, then the whole tract is: 120.6.4 = 720.4 = 2880 hectaresLets take that every one of the four tractors for the 5 days ploughed x hectares. Therefore the finished work:
5.4. x = 20 . x hectares and this is the whole tract  2880 hectares
So we get 20x = 2880 <=> x = 2880/20 = 144 decares a day ploughed every on of the four tractor-drivers.

Problem 5
One student thought of a number, multiplied it by 2. From the received product subtracted 138 and got 102 . Which is the number the student thought of?
Solution:
Lets take the thought number as x, when he multiplied it with 2 he got (2x); from which he subtracted 138 i.e. 2*x - 138 and by condition received 102 <=> 2. x -138 = 102 We must solve this equation to find the thought number
2*x - 138 = 102 <=> 2x = 240 <=> x = 240/2 <=> x = 120

Problem 6
I thought of a number, divided it by 5, from the received quotient I subtracted 154 and got 6. Which is the number i thought of?
Instruction: The thought number is x and the equation: x/5 -154 = 6
Solve the equation by yourself. Answer x = 800

Problem 7
The distance between two towns is 380km. A car and a lorry started from the two towns at the same time. At what speed drove the two vehicles, if the speed of the car is with 5km/h faster than the speed of the lorry and we know that they met after 4 hours?
Solution:
The basic dependence that is used in problems with movement is that the distance is equal to the speed multiplied by the time S = V.t

 V km./h. t h. S km. Car x + 5 4 4(x +5) Lorry X 4 4x

4(x + 5) + 4x = 380 <=> 4x + 4x = 380 - 20 <=> 8x = 360 <=> x = 360/8 <=> x = 45
Therefore the lorry was driving with 45 km/h., and the car with 50 km/h.

Problem 8
One of the sides of a rectangle is with 3cm. shorter than the other one. Find the sides of the rectangle if we know that, if we increase every side with 1cm., the surface of the rectangle will be increased with 18cm2
Solution:
Lets take that one of the sides is x cm. (x > 3), then the other will be x  3 cm. For the surface we find S1 = x(x - 3) cm2. If we increase the proportions with 1cm. the sides will be (x + 1) cm. and (x - 3 + 1 ) = (x - 2) cm. and these are the new proportions of the rectangle so the surface is S2 = (x + 1).(x - 2) cm2 and by condition it is with 18 cm2 bigger than the first one. Therefore we get the following equation:
S1 +18 = S2 <=> x(x - 3) + 18 = (x + 1)(x - 2) <=> x2 - 3x + 18 = x2 + x - 2x - 2 <=> 2x = 20 <=> x = 10 And so the sides of the rectangle are 10 cm. and (10 - 3) = 7cm.

Problem 9
For one year from two cows were milked 8100l. The next year the first cow increased her yield of milk with 15% and the second one with 10%. So they milked 9100 l. from the two cows. How many litres are milked from every cow during the first and the second year?
Solution:
If during the first year the first cow gave x l., then the second one gave (8100  x) l. The increase in the yield of milk is 15% of x, i.e. 15/100.x and 10% of (8100  x), i.e. 10/100 . (8100  x) . Then during the second year the two cows gave the amount milk from the first year + the increase of the second year
So we get the following equation: 8100 + 15/100.x + 10/100 . (8100  x) = 9100
Therefore 8100 + 3/20x + 1/10 (8100  x) = 9100 <=> 1/20 . x = 190 <=> x = 3800
And so for the first year the milked 3800 and 4300 l. from every cow and for the second year 4370 l and 4730 l.

Problem 10
The distance between stations A and B is 148 km. From station A to station B leaves an express train which proceeds with 80 km/h and at the same time from station B towards station A leaves a goods train with 36 km/h. We know that before the two trains meet at station C the express train made a 10min and the goods train - 5min. Find:
a) The distance between station C and station B
b) At what time the goods train left station B if the meeting with the express train at station C was at 12 oclock.
Solution
a) We mark the distance from station B to station C with x km. 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 ran (148 x)/80 + 10/60 hours and the goods train x/36 +5/60. Because the trains left at the same time these times are equal : (148  x)/80 + 1/6 = x/36 + 1/12 We reduce to a common denominator, which for 6, 12, 36, 80 is 720 We release from denominator and we get:
9(148  x) +120 = 20x +60 <=> 1332  9x + 120 = 20x + 60
<=> 29x = 1392 <=> x = 48 Therefore the distance from station B to station C is 48 km.
b) By the time of meeting at station C the goods train ran 48/36 + 5/60 hours, i.e. 1 hour and 25 min.
Therefore he left station B in 12 - 1.25/60 = 10.35/60 oclock, i.e. in 10 h. and 35min.

Problem 11
A motorman should have taken a distance from town A to town B for exact time. Two hours after he left, he noticed that he covered 80 km and if he keeps that speed he will arrive in B with 15 min delay. So he increased the speed with 10km/h and arrived in town B 36 minutes earlier. Find:
a) The distance between the two towns;
b) The exact time that the motorman should have taken the distance from A to B
Solution:
We mark the distance from A to B with x km. Because the motorman took 80km for 2 hours his speed is V = 80/2 = 40 km/h. With that speed he would have taken the whole distance for x/40 h, delaying with 15min, i.e. the exact time is x/40  15/60 h. The rest of the distance (x - 80) km. he took with V = 40 + 10 = 50 km/h.
So the time he took the distance from A to B, is 2 +(x - 80)/50 h. and it is with 36 min. earlier than expected. Therefore the expected time is 2 + (x -80)/50 + 36/60 When we equalize the expressions for the expected time, we get the equation:
x/40  15/60 = 2 + (x -80)/50 + 36/60 <=> (x - 10)/40 = (100 + x - 80 + 30)/50 <=> (x - 10)/4 = (x +50)/5 <=> 5x - 50 = 4x + 200 <=> x = 250
So the searched distance is 250 km. The exact time we will find by substituting x with 250 in of the sides of the first equation, for example;
x/40  15/60 = 250/40  1/4 = 25/4  1/4 = 24/4 = 6 hours

Problem 12
To be able to make one order for pieces in time a team should make 25 pieces a day. After 3 days the team increased the productivity with 5 pieces and made 100 pieces over the plan for the exact time. Fine how many details the team made and for how many days?
Solution:
Lets take the days the team worked as x. Then 25.x are the pieces that they should have made. With the new rate they made:
1.25 + (x - 3)(25 + 5) = 75 + 30.(x - 3) and they are with 100 more than expected.
Then: 25. x = 75 + 30(x -3)  100 <=> 25x = 75 +30x -90  100 <=>
190 -75 = 30x -25 <=> 115 = 5x <=> x = 23
So the days are 23 and the made pieces are 23.25 = 575

Problem 13
There are 24 students in 7a class. During a youth brigade they planted a total of 24 birches and roses where every girl planted 3 roses each and every three boys planted 1birch. Find how many birches and roses are planted from the students of 7a?
Solution:
Lets take the number of planted roses as x, then the birches are (24  x) . If every girl planted 3 roses each the number of girls is x/3 . From the fact that 3 boys planted one birch follows that the boys are 3(24 - x).
The total number of students in this class is 24, i.e. x/3 + 3(24  x) = 24 <=> x + 9(24  x) = 3.24 <=> x +216  9x = 72 <=> 216  72 = 8x <=> 144/8 = x <=> x = 18
Therefore the planted roses are 18 and the birches are 24  x = 24 - 18 = 6.

Problem 14
From town A went a car, by exact road, to town B at speed V = 32km/15/60 = x - 0,25. After 3 hours from the departure the driver made a 15min stop in town C. Because of some damage on the road he changed the road to town B with another one, which was with 28km. longer than the exact one and he went at V = 40km/h. If the car has arrived with 30min delay in town B, find:
a) The distance the car has covered
b) The time that took the driver to get from C to B
Solution:
From the condition of the problem we dont know if the 15min stop in town C is expected or it is made because of the road damage. So we will observe both cases.
1st case . If the stop is expected and when he went directly to B. For both cases we will observe only the movement from C to B. The real movement (by the longer road) we will take as x h.
Then the covered distance from C to B is S = 40.x km. The time from C to B if taking the exact road is x - 30/60 = x - 1/2h.The distance that he should have covered from C to B if there was no road damage is (x - 1/2).32km, which is with 28 km.shorter than 40.x km. So the equation we get is
(x - 1/2).32 + 28 = 40x <=> 32x -16 +28 = 40x <=> 8x = 12 <=> x = 12/8 x = 1.4/12 = 1.20/60 = 1h.20min.
So the car took the distance from C to B for 1hour and 20 min.
And the covered distance from A to B is 3.32 + 12/8.40 = 96 + 60 = 156 km.
II solution Lets take that the 15min stop is done only in the real case, i.e. because of the necessary taking of the longer road. Lets again the movement which practically is made from C to B, is for x hours. Then the distance is again S = 40.x km. The exact movement from C to B the time is x - 30/60 - 15/60 = x -45/60 = x - 3/4 h. The exact distance from C to B is 32(x - 3/4)km. and it is 28 km. shorter than 40.x, i.e.
32(x - 3/4) + 28 = 40x <=> 32x - 24 +28 = 40x <=> 4 = 8x <=> x = 1/2hours * x = 30 min. Then the time for real movement from C to B is 30min. The covered distance is 3.32 + 1/2.40 = 96 + 20 = 116 km.

Problem 15
To be able to plough a tract in time must plough 120hectaresa day. For technical reasons he ploughed 85hectares a day, because of that he ploughed 2 days more than the exact time and 40hectares left to be ploughed. Find how many decares is the whole tract and how many days was the exact time to be ploughed?
Solution:
Lets take the days the tract should have been ploughed as x Then the whole tract is 12.x hectares If taking the real ploughing the time is x + 2 or 85 hectaresa day, therefore it was ploughed 85(x + 2), which is with 40 hectares less than the whole tract. The equation is:
120. x = 85(x + 2) + 40 <=> 35x = 210 <=> x = 6 So the days the tract should have been ploughed are 6 and the tract is 120.6 = 720 hectares

Problem 16
For 24 days a turner makes exact quantity of pieces. By increasing his daily production with 5 pieces he worked 22 days and made 80 pieces over the exact quantity. Find the daily production and how many pieces should have made?
Solution:
Lers x pieces is his daily production. For 24 days he will make 24.x pieces. His new production is x + 5 pieces and for 22 days he will make 22.(x + 5) details, which are with 80 more than 24x. Then the equation is:
24. x + 80 = 22.(x +5) <=> 30 = 2x <=> x = 15
His daily production is 15 pieces and totally he should have made 15.24 = 360 pieces.

Problem 17
A motorman took half of the distance between two towns for 2h.30min. and after that he increased his speed with 2km/h.He took the second part of the distance for 2h.20min. Find the distance between the two towns and the original speed of the motorman?
Solution:
If on the first half of the distance the speed is x km/h, in the second one it will be x + 2 km/h.The distances taken with speed 2.30/60.x km and 2.20/60.(x + 2)km and by condition they are equal. From the equation: 2.30/60.x = 2.20/60.(x +2) we get x = 28km/h
For the distance between the two towns we find 2.2.20/60.28 = 140 km.

Problem 18
A train, after taking half of the distance between two stations A and B with 48km/h, made a 15min stop. After that he increased his speed with 5/3 m/sec. and arrived on time in station B. Find the distance between the two stations and the speed of the train after the stop?
Solution:
First we will determine the speed of the train after the stop. The incensement of 5/3m/sec = 5*60*60/3*1000 km/h = 6km/h Then the new speed is 48 + 6 = 54 km/h. If the first half of the distance is taken for x hours, the second one for x  15/60 = x - 0.25h
Then the equation is: 48*x = 54*(x - 0.25), from where x = 13.5 h. The searched distance is determined from 2*48*13.5 = 216.69 km.

Problem 19
A worker can finish exact work for 15 days, other worker can finish only 75% of that work for the same time. At first the second worker worked several days and then the first one joined him and together they finished the rest of the work for 6 days.
Find how many days worked every worker and what percent of the work has done each one of them?
Solution:
First we will find the daily production of every worker. If we take the whole work as unit(1), the production of the first one is 1/15 and the production of the second is 75% of 1/15, i.e.
75/100.1/15 = 1/20 Lets take that the second worker worked alone x days. Then the work he finished will be x/20. For the 6 days work done in common they finished 6.(1/15 +1/20) = 6.7/20 = 7/10
The sum of x/20 and 7/10 gives the whole work, i.e. 1. So we get the equation:
x/20 +7/10 = 1 <=> x = 6 The second worker worked 6 +6 = 12 days and the first one only 6 days. The work finished from the second worker is 12.1/20 = 60/100 = 60%, and from the first one 6.1/15 = 40/100 = 40%

Problem 20
Tractor-drivers planned to plough one tract by ploughing 120hectaresa day. After the first two days they increased the daily production with 25% and that is why they finished two days before the exact day. Find:
a) How many decares is the whole tract?
b) How many days took to plough the whole tract?
c) How many days would have taken to plough to whole tract if following the exact plan?
Solution:
First of all we will find the new daily production of the tractor-drivers in decares: 25% of 120 decares are 25/100.120 = 30 dec, therefore 120 + 30 = 150 hectares is the new daily production. Lets take the initial needed time for ploughing as x days. Then the tract is 120.x hectares The same tract can be found when to 120.2dec is added 150.(x -4)hectares Then the equation is
120x = 120.2 + 150.(x -4) <=> x = 12 So 12 days were needed if following the plan but actually the tract was ploughed for 12 -2 =10 days. The tract is 120.12 = 1440 hectares

Problem 21
To cut down a tract of grass in exact time, a team of mowers should plough 15hectares daily. The first 4 days they worked like this and then increased the daily production with 33.1/3%.They finished the work 1 day earlier. Find:
A) how many deares is the whole tract?
B) How many days took to cut the whole tract?
C) How many days would have taken to cut the whole tract if following the exact plan?
Instruction: See problem 20 and solve by yourself
Answer: A) 120 dec B) 7 days C) 8 days

Problem 22
A train should have taken the distance from A to B according to the schedule for exact time. If the train leaves station A and proceeds with 75km/h he will arrive in station B 48 minutes earlier. If he proceeds with 50km/h for this time he will arrive 40km prior to station B. Find:
A) the distance between the two stations;
B) the time the train takes the distance according to the schedule;
C) the needed speed to keep to the schedule;
Solution:
Lets take the time for movement from A to B as x hours. Then the distance from A to B can be found in two ways. First 75(x - 48/60)km., and second, 50x + 40 km. So we get the equation:
75(x - 48/60) = 50x + 40 <=> x = 4 is the time by schedule. The distance between the two stations is 50.4 +40 = 240 km. Then the speed he needs to keep to the schedule is 240/4 = 60 km/h

Problem 23
From two towns A and B, with distance 300km between them, at the same time left two trains. We know that the speed of one of the trains is with 10km/h faster than the speed of the other one. Find the speed of the two trains if 2 hours after their departure the distance between them is 40km.
Solution:
Lets take the speed of the slower train as x km/h.The speed of the other will be x + 10 km/h. After 2 hours they will cross 2x km and 2 9x +10) km.Then the whole distance from A to B is 2x + 2(x +10) +40 = 4x +60 km, if the trains hasnt already met or 2x +2( x +10) -40 = 4x -20 km, if they have met. So we get the following two equations:
4x + 60 = 300 <=> 4x = 240 <=> x = 60 or
4x  20 = 300 <=> 4x = 320 <=> x = 80
So the speed of the slower train is 60 km/h or 80 km/h and the speed of the other one is 70 km/h or 90 km/h

Problem 24
A bus takes the distance between two towns A and B for exact time. If the bus goes with 50km/h he will arrive in B with 42min delay and if he increases his speed with 5.5/9 m/sec., he will arrive in B 30min before the exact time. Find:
A) the distance between the two towns;
B) the exact time for the bus to take the distance;
C) the speed of the bus(by schedule) for the exact time.
Solution:
First we will determine the new speed of the bus. The incensement is 5.5/9 m/sec. = 50/9 m/sec = 50.60.60/9.1000 km/h = 20 km/h Therefore the new speed is V = 50 +20 = 70 km/h If by schedule the time for movement is x hours, at speed 50 km/h he moved from A to B for x +42/60 h, when V = 70km/h for x  30/60h Then
50(x +42/60) = 70(x -30/60) <=> 5(x +7/10) = 7(x -1/2)
<=> 7/2 +7/2 = 7x -5x <=> 2x = 7 <=> x = 7/2
So the time by schedule is 3h.30min.
The distance from A to B is 70(7/2 -1/2) = 70.3 = 210 km and the speed by schedule 210/(7/2) = 60km/h.

Send us math lessons, lectures, tests to: