maths word problems and solutions

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 = 120Therefore 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 4One 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 hectaresSo we get 20x = 2880 <=> x = 2880/20 = 144 decares a day ploughed every on of the four tractor-drivers.

Problem 5One 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 number2*x - 138 = 102 <=> 2x = 240 <=> x = 240/2 <=> x = 120

Problem 6I 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 = 6Solve the equation by yourself. Answer x = 800

Problem 7The 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 = 45Therefore the lorry was driving with 45 km/h., and the car with 50 km/h.

Problem 8One 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 9For 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 yearSo we get the following equation: 8100 + 15/100.x + 10/100 . (8100 – x) = 9100Therefore 8100 + 3/20x + 1/10 (8100 – x) = 9100 <=> 1/20 . x = 190 <=> x = 3800And 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 10The 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 Bb) At what time the goods train left station B if the meeting with the express train at station C was at 12 o’clock.Solutiona) 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 11A 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 = 250So 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 12To 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 = 23So the days are 23 and the made pieces are 23.25 = 575

Problem 13There 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 14From 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 coveredb) The time that took the driver to get from C to BSolution: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 15To 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 16For 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 = 15His daily production is 15 pieces and totally he should have made 15.24 = 360 pieces.

Problem 17A 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/hFor the distance between the two towns we find 2.2.20/60.28 = 140 km.

Problem 18A 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 - 0.25hThen 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 19A 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 = 7/10The 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 20Tractor-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 is120x = 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 21To 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 yourselfAnswer: A) 120 dec B) 7 days C) 8 days

Problem 22A 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 23From 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 or4x – 20 = 300 <=> 4x = 320 <=> x = 80So 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 24A 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 = 7/2So 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.

Friday, October 21, 2011

Age Problem 8 Solution

This is the solution to the algebra age problem asked by Joshua that read, "A man is 4 years older than his wife and 5 times as old as his son. When the son was born, the age of the wife was six-sevenths that of her husband's age. Find the age of each."Ok so here we go...

"A man is 4 years older than his wife"m = w + 4Now we will subtract 4 from both sides to use this conveniently later ( you will see)w = (m-4)

"and 5 times as old as his son."m = 5s

"When the son was born, the age of the wife was six-sevenths that of her husband's age."w - s = 6/7(m-s)

7(w-s) = 6(m-s) -->7w - 7s = 6m - 6s --->7w = 6m - 6s + 7s --->7w = 6m + sreplace w with (m-4)7(m-4) = 6m + s7m - 28 = 6m + s7m - 6m = s + 28m = s + 28replace m with 5s5s = s + 285s - s = 284s = 28s = 7 yrs is the son's agethen5(7) = 35 yrs is Dad's ageand35-4 = 31 yrs is Mom's age

Now check work..."When the son was born, the age of the wife was six-sevenths that of her husband's age."31 - 7 = 6/7(35 - 7)24 = 6/7*2824 = 24;

http://algebra-word-problem-solutions.blogspot.com/2011/10/age-problem-8-solution.html

Posted by Klitschko Fan at 5:21 PM 1 comments

Labels: Age Problem Solutions

Sunday, October 9, 2011


This is the solution to a algebra age word problem as asked by a anonymous user: "Nona is one-third as old as her mother. Five years ago, she was only one-fifth of the age of her mother. How old is Nona now? "

It is easier to set this up with two unknowns say x and y and have 2 equations.

The 1st equation would be: x = 1/3(y) or x = y/3 since Nona is currently 1/3 as old as her mother. Now 5 years ago she was only 1/5th the mothers age so the other equation would be..

x -5 = 1/5(y - 5) multiply by 5..

5x - 25 = y - 5 so

y = 5x - 20 .. plug this back into the 1st equation and you get..

x = (5x - 20)/3 - -multiply by 3

3x = 5x - 20 so 2x = 20 and x = 10

So Nona is 10 years old currently and the mom is 30. To check this go 5 years back... Nona would be 5 and the mother would be 25 ------ 5/25 = 1/5ths of her mothers age...

Posted by Klitschko Fan at 9:41 AM 0 comments


Tuesday, December 14, 2010


This is the answer to the age algebra word problem that asked, "Bob is one third the age of his father. In 12 years he will be half the age of his father. How old is each now?"


http://algebra-word-problem-solutions.blogspot.com/search/label/Age%20Problem%20Solutions

http://algebra-word-problem-solutions.blogspot.com/2011/10/age-problem-7-solution.html#comment-form






http://www.addthis.com/bookmark.php?pub=klitschkofan&url=http://algebra-word-problem-solutions.blogspot.com/2011/10/age-problem-8-solution.html&title=Age%20Problem%208%20Solution


Ok Bob being 1/3 the age of his father is the same thing as saying the father is 3 times as old as him. It makes it easier to work with. So let Bob be x and his father be 3x. And in 12 years bob will be half as old as his father. So we have..

x + 12 = (3x + 12) / 2

2x + 24 = 3x + 12

x = 12

So Bob is 12 years old and his father is 3(12) or 36 years old now. In 12 years Bob will be 12 + 12 or 24 years old and his father will be 36 + 12 or 48 years old which means Bob will be half as old so it checks out.



Thursday, November 25, 2010


This is the algebra age word problem that read Pol is 10 years younger than greg. In 7 years, he will be 10 years more than one halfas old as greg. Find their age at present. help me solve it.

Let Greg be X and Pol be X - 10 as far as current ages. In 7 years (thats x + 7) he will be 10 years more than 1/2 as old as greg (thats (x - 10 + 10)/2 or x/2). So set them equal..

X and x -10

x + 7 and x - 10 + 7 28 and 25

(x+7)/2 + 10 = x - 3

x + 17 = 2x - 6 ******** Correction 2 * 10 = 20 + 7 = 27 So it should be

x + 27 = 2x - 6

x =33 ---> Which is Greg's age so Pol's is x-10 or 33 -10 = 23

So Pol is 23 and Greg is 33








Sunday, August 2, 2009


This is the solution to the algebra age problem # 4... "A man is 27 years older than his son and 10 years from now, he will be twice as old as his son. how old is each now?"

Let x = the son and x + 27 = the man(father)

in 10 years the son will be x + 10 and the father x + 27 + 10 or x + 37 years old. So if he will be twice as old as his son then "twice" the sons age then will "equal" each other.

So, 2(x+10) = x + 37multiply by 2 on the left side

2x + 20 = x + 37

x = 17 This makes sense since 10 + 17 = 27 and 17 + 37 = 54 which is twice the sons age 10 years from now. So the son is 17 and the father is 17 + 27 or 44 years old.

Tags: Age Problem Solutions, Algebra Word Problem Solutions



Thursday, March 20, 2008


Greg will be x and Greg's father will be x + 30. In 15 years Greg will be x + 15 and Greg's father will be x + 30 + 15 or x + 45, and the sum of their ages will be 130.

So,

x + 15 + x + 45 = 130





http://technorati.com/tag/Algebra+Word+Problem+Solutions

http://technorati.com/tag/Age+Problem+Solutions







2x + 60 = 130

2x = 130 - 60

2x = 70

x = 70/2

x = 35 which is Greg's current age. Greg's father is 35 + 30 or 65 years old. Just to check add 15 years or 30 total to their ages. 35 + 65 + 30 = 130



Sunday, March 16, 2008


Here's a non-algebraic approach to the problem:

If Bobs current age were 8 then 8 years ago he would be zero, so he has to be older than 8 or at least 9 years now. So assume 9 and 36 -- 8 years ago would make 1 and 28 which is 28 times and not what were looking for.. 10 and 40 ---- 8 years ago would make 2 and 32 which is 16 times not what were looking for... try 11 and 44-- 8 years ago would make 3 and 36-- bingo 12 times- So, Jason is 44 and Bob is 11 years old.

Algebraically we have:

x for Bobs age and

4x for Jason

So 4x = x

But 8 years ago Jason was 12 times older so

4x - 8 = 12(x-8)

4x - 8 = 12x - 96

88 = 8x

x = 11 So Bob is currently 11 years old and Jason is 4(11) or 44 years old.








Saturday, March 15, 2008


For a non-algebraic approach:

If Steves current age was 20 then 20 years ago he would be zero, so he has to be at least 21 years old. This would make Bob 42 and Steve 21. 20 years ago they would be 22 and 1, which is 22 times as old and not 6 times. Were not that far off though. So go up by increments. Try Steve at 22 and Bob 44-- 20 years ago would be 2 and 24 which is 12 times-- at 23 and 46 we can see than 20 years ago they would be 3 and 26 which is over 8 times were close-- and at 24 and 48 20 years ago would make 4 and 28 which is 7 times and really close.. at 25 and 50 20 years ago we would have 5 and 30 which IS 6 times older and what were looking for. So, Steve is 25 and Bob is 50...

For the algebraic approach we know that if something is twice as old, big, fast, etc as something else it is 2x more than x. So in this problem Bob is 2x as old as x Steve is.

Twenty years ago would be :

2x - 20 for Bob

and

x - 20 for Steve

but, 20 years ago Bob was 6 times older so

2x - 20 = 6(x - 20)

2x - 20 = 6x - 120

4x = 100

x = 100/4

x = 25

So x or Steve is 25 and Bob is twice or 2(25): 50 years old.






y, December 3, 2011

Finance Problem 4 Solution

This is the solution to the Algebra finance word problem a anonymous user asked, "An administrative assistant orders cellular phones for people in her department. The brand A phones cost $89.95 and the brand b phones $34.95.If she ordered 3 times as many brand b phones as brand a phones at a total cost of $584.40, how many of each did she order?"

Ok to make things easier lets make it cents. Just remember to divide by 100 when done to get the answer in dollars and cents...

So brand A is 8995 and b is 3495 which had 3 times as many sold as a and all the combinations of phones totalled 58440.

So 8995x + 3495(3x) = 58440

8995x + 10485x = 58440

19480x = 58440

x = 3 So there were 3 brand A's and 3(3) = 9 brand B's sold.


Labels: Finance Problem Solutions


This is the solution to the Algebra finance word problem asked by an anonymous user which read, "Alex has made 42 of the 48 payments he owes on his car but is having trouble continuing to make the payment in full. He has made an arrangement with the bank to pay 2/3rds of his monthly payment, rather than the entire payment, every month. How many months will it take Alex to pay off his car loan with this new payment arrangement?"

Ok, 1st find the remaining months : 48 - 42 = 6 So there are 6 months at full payments. But, you are no longer doing full payments you are doing 2/3s of a payment. So 2/3 of what would equal out 6 ? Or

2x/3 = 6

2x = 18

http://algebra-word-problem-solutions.blogspot.com/2011/12/finance-problem-3-solution.html

http://algebra-word-problem-solutions.blogspot.com/search/label/Finance%20Problem%20Solutions

http://algebra-word-problem-solutions.blogspot.com/2011/12/finance-problem-4-solution.html#comment-form




http://www.addthis.com/bookmark.php?pub=klitschkofan&url=http://algebra-word-problem-solutions.blogspot.com/2011/12/finance-problem-4-solution.html&title=Finance%20Problem%204%20Solution

x = 18/2 = 9 .. So it will take Alex 9 months to pay off his car loan with the new payment arrangement..



This is the finance problem 2 solution which read: "A local furniture store is selling all $850 mattresses at 35% off. Alyssa is buying a mattress and has a coupon for an additional 10% off the sale price. What will Alyssa pay for her mattress? "

Ok you just have to take the discount off twice. 35% off of $850 would be $297.50 so 850 - 297.50 = $552.50 is the sale price. Now she has a coupon for another 10% off that sales price. So 10% of 552.50 is 55.25 and 552.50 - 55.25 = 497.25

So Alyssa will pay $497.25 for her mattress....



Saturday, January 23, 2010


This is the solution to the algebra finance word problem that An anonymous poster recently asked which read:

"An employee's new salary is $19,110 after getting a 5% raise. What was the salary before the increase in pay?"

Ok easiest way to think about this is try something you do know for sure. Whats 5% of 100? 5 right--- Ok, so what if the problem said an employee's new salary was $105 after a 5% increase in pay what was the salary before... Ok you already know the answer 100.... So how would u come up with the new total? Well 5% of 100 is 5 and then you add it to the original amount of 100 which makes 105. So, you set up your problem like that:

5% can be written 5/100

so,

5/100 * x + x = 105 multiply by 100



http://algebra-word-problem-solutions.blogspot.com/2010/11/finance-problem-2-solution.html#comment-form






5 * x + 100x = 10500

5x + 100x = 10500

105x = 10500x = 10500/105 = 100 so 100 is the answer you were looking for and it checks out..

You solve this problem exactly the same way.

5/100 * x + x = 19110 multiply by 100

5x + 100x = 1911000

105x = 1911000

1911000/105 = x

x = 18200 So check whats 5% of 18200 ? its 910 Now add 910 to 18200 and you get 19110. So it checks out.

Coin Problem 6 Solution

This is the coin problem 6 solution which read "a stack of pennies and dimes has a total value of $2.31. how many dimes are in the stack if there are twice as many dimes as pennies?"

make $2.31 into cents so 231 cents .. a dime is worth 10x and penny is worth x. So if you had 1 dime and 1 penny you would have 10(1) or 10 cents plus 1(1) or 1 cent which is 11 cents. Were trying to have it add up to 231 cents though and there are twice as many dimes 2(10x) or 20x as pennies.

So we have 20x + x = 231

21x = 231

x = 11 So there are 11 penny's or 11 cents and 22 dimes or 220 cents .. 220 cents plus 11 cents equals 231 cents or $2.31.


Labels: Coin Problem Solutions

http://algebra-word-problem-solutions.blogspot.com/search/label/Coin%20Problem%20Solutions

http://algebra-word-problem-solutions.blogspot.com/2010/11/coin-problem-6-solution.html#comment-form

http://algebra-word-problem-solutions.blogspot.com/2010/11/coin-problem-6-solution.html


http://www.addthis.com/bookmark.php?pub=klitschkofan&url=http://algebra-word-problem-solutions.blogspot.com/2010/11/coin-problem-6-solution.html&title=Coin%20Problem%206%20Solution

Wednesday, March 12, 2008


Let the total amount in cents = 2251

let x = the # pennies and the value = 1(x) or just x

let 2x = the # of quarters and 25(2x) or 50x = the value in quarters

let 20(2x) or 40x equal the # of 50 cent pieces and 50(40x) or 2000x = the value of 50 cent pieces

So, we have:

x + 50x + 2000x = 20512051x = 2051x = 2051/2051x = 1

so we have (1) pennies

2(1) or 2 quarters

and 40(1) or 40 half dollars



Tuesday, March 11, 2008


19.43 is 1943 cents.

In this problem, it helps to think about what the easiest setup would be before just setting x equal to the 1st thing you see. If you let x = pennies, well that would work, but you would end up worth a lot of fractions that you could avoid if you started with x = quarters instead.

So let x = the # of quarters and 25x the value

2x = the number of dimes and 10(2x) = the value or 20x

4x = the # of pennies and 1(4x) or 4x the value of pennies







4x + 1 = the # nickels and 5(4x+1) or 20x + 5 the value of nickels

and

4(4x+1) or 16x + 4 the # of half dollars and 50(16x + 4) or 800x + 200 the value

So altogether we have 25x + 20x + 4x + 20x + 5 + 800x + 200 = 1943

869x + 205 = 1943

869x = 1738

x = 1738/869

x = 2

So we have 2 quarters, (50 cents)

2(2) or 4 dimes (40 cents)

4(2) or 8 pennies (8 cents)

4(2) + 1 or 9 nickels (45 cents)

and 16(2) + 4 or 36 half dollars (1800 cents)

just to check if you add those cents up it will equal 1943 cents which is $19.43



Monday, March 10, 2008


Again make $10.35 equal to cents --- 1035 cents 1st.

Algebraically we have

x = 32x 32 cent stamps

x + 2 = 25(x + 2) 25 cent stamps and






2(x + 2) or 2x + 4 = 50(2x + 4) 50 cent stamps

So, we then have..

32x + 25(x + 2) + 50(2x + 4) = 1035

32x + 25x + 50 + 100x + 200 = 1035

157x + 250 = 1035

157x = 1035 - 250

157x = 785

x = 785/157

x = 5

So we have 5 32 cent stamps, 5 + 2 or 7 25 cent stamps, and 2(5 + 2) or 14 50 cent stamps...




1st of all on all these problems convert the total amount of money from dollar to cents-- $100 is equal to 10000 cents (just add 2 zeros), $7 is equal to 700 cents, and you already have 37 cents, so altogether you have 10000 + 600 + 87 = 10737 cents.

From the problem you can see that everything is relative to pennies so let that be 1(x) or just x

1 more than twice as many nickels as pennies is (2x + 1) and and since its nickels 5(2x +1) or 10x + 5

twice as many dimes as nickels 2(2x +1) = 4x + 2 and since its dimes 10(4x + 2) or 40x + 20

1 more than twice as many 50 cents pieces than dimes (this is tricky wording since you may expect the denomination to go to a quarter) which is 2(4x + 2) + 1 or 8x + 5 and since its 50 cent pieces 50(8x + 5) or 400x + 250

and twice as many quarters as 50 cent pieces or 2(8x + 5) or 16x + 10 and since its quarters 25(16x + 10) or 400x + 250

so algebraically we have






x + 10x + 5 + 40x + 20 + 400x + 250 + 400x + 250 = 10737 =851x + 525 = 10737 = 851x = 10737 - 525 =851x = 10212 =x = 10212/851

x = 12

So you have 12 pennies, 2(12) + 1 or 25 nickels, 4(12) + 2 or 50 dimes, 8(12) + 5 or 101 Fifty cent pieces, and 16(12) + 10 or 202 quarters....

Yeah, a little tedious arithmetic wise...




On any kind of coin problem its important to turn any dollar amount (if given in dollars) to cents and to make sure you give the appropriate value of each denomination ---- a penny 1 or -- nickel 5--- dimes 10-- quarter 25-- sounds kind of obvious but a lot of problems are taken care of if you just know that information for sure.

Alright if you have 10 pennies you have 10(1) == 10 cents So,

If you have 12 nickels you would have 12(5) = 60 cents

If you had 5 dimes you would have 5(10) = 50 cents

and if you had 10 quarters you would have 10(25) = 250 cents or $2.50

To solve this problem you 1st set up the information given. You can see that everything is relative to the pennies (1 more nickel than pennies and 6 times as many dimes as pennies) So let pennies equal 1x or just x since 1 times anything is just the number...

you have x + 1 nickels at 5 cents so 5(x+1) nickels

and 6 times as many dimes as pennies or 10(6x) or 60x

Theres no need to change the dollars to cents as the cents are already given 71 cents.

So the equation would be

x + 5(x+1) + 10(6x) = 71 =






x + 5x + 5 + 60x = 71 =

66x = 66 =x = 66/66 = 1

So thers 1 penny, (1 + 1) or 2 nickels and 6(1) 6 dimes

To check if you add those up 1 + 10 + 60 you get 71 so the answer is correct.

Thursday, October 9, 2008

Lever Problem 4 Solution

In this Algebra Lever problem solution, you again will have a problem where the distance times the weights have to equal on both sides. Also since Steve sits twice as far, he is 8 feet away since Samuel is 4 feet away and Steve sits twice as far as Samuel.

Here's the algebra equation:

65(4) + 85(8) = 200x

260 + 680 = 200x

940 = 200x

040/200 = x

94/20 = x = 4 and 14/20 or 4 and 7/10 or 4.7 feet is how far Big Billy would have to sit on the opposite side of the teeter totter across from Samuel and Steve.


Labels: Lever Problem Solutions

Thursday, June 12, 2008


1st piece of information that's important in the Algebra lever problem is that the teeter totter was 12 feet long and the balance point was directly in the middle(which most teeter totter's are). So each side is 6 feet. This means if Bob and Susan sat on opposite ends then they were 6 feet from the fulcrum. They weigh 70 and 50 lbs respectively. On the same side as Susan 5 feet away from the center 40 lb Chrissy sat. So we got the weights and distances for everyone but Steve who weighs 35 lbs but is an unknown (x) distance from the center on the same side as Bob.

http://algebra-word-problem-solutions.blogspot.com/2008/06/lever-problem-3-solution.html

http://algebra-word-problem-solutions.blogspot.com/search/label/Lever%20Problem%20Solutions

http://algebra-word-problem-solutions.blogspot.com/2008/10/lever-problem-4-solution.html#comment-form



http://www.addthis.com/bookmark.php?pub=klitschkofan&url=http://algebra-word-problem-solutions.blogspot.com/2008/10/lever-problem-4-solution.html&title=Lever%20Problem%204%20Solution

As we said before the distance times the weight of one side has to equal the other. In this case since there's more than 1 person on each side we add the respective weights and distance and set them equal to one another...

This side:

Susan = 50(6) Chrissy = 40(5)

is equal to this side:

Bob = 70(6)Steve = 35(x)

300 + 200 = 420 + 35x500 = 420 + 35x

80 = 35x x = 80/352 & 10/35 feet or 80/35 from the center is where Steve is located on the same side as Bob.

Just to check 500 = 420 + 35(80/35) = 420 + 80 == correct



Monday, May 12, 2008


In lever problems you are trying to get the weight * the distance of one side to equal the weight times the distance of the other. The fulcrum would be considered the balance point, and since a weight of 100 lbs is on that side we can see that the weight of 100 * 2 = 200 has to equal the distance of 8 (10 - 2) times a certain amount of weight which were trying to figure out for the other side. So algebraically we have:

100 * 2 = 8 * x

200 = 8x

x = 200/8

x = 25






This means that a weight of 25 lbs would have to be applied to one side of the lever to lift the 100 lbs. As far as levers are concerned this makes sense. You definitely want to exert less force on one end that the weight of object your trying to move!





With any type of lever problem it's important to realize that the weight of object 1 TIMES the distance of that object from the center(or fulcrum) is going EQUAL the weight of object 2 TIMES the distance of that object from the center(fulcrum).

So in this problem you know the to weights are Sam at 100 lbs and James at 150 lbs, and the distance from the fulcrum is known for Sam which is 8(feet). However, you don't know James' distance so thats what were trying to find.

So set them equal

100 * 8 = 150 * x

800 = 150x

x = 800/150

x = 5 1/3 (feet) or 5 feet 4 inches (since 1(ft)/3 * 12 inches/ft = 4 inches)

Mixture Problem 7 Solution

This is the solution to a new algebra mixture problem as asked by Califax who asked, "A 50-lb solution of acid and water is 20% (by weight). How much pure acid must be added to this solution to make it 30% acid?"

1/5 = 20 % ..... 3/10 = 30 % x = the unknown amount in lbs we want to add of 100% or 1/1 or 1 or just x of pure acid.

50(1/5) + x = 3/10 (50 + x) multiply by 10...

100 + 10x = 3(50 + x)

http://algebra-word-problem-solutions.blogspot.com/2011/10/mixture-problem-7-solution.html






100 + 10x = 150 + 3x

7x = 50 = 50/7 = 7 and 1/7th lbs of pure acid needs to be added to make the solution 30% acid.


Labels: Mixture Problem Solutions

Wednesday, October 7, 2009

Mixture Problem # 6 Solution

This is the solution to the algebra mixture word problem # 7 which read:

"If a merchant has two types of tea, one worth $2.70 per kilogram and the other worth $3.00 per kilogram, how many kilograms of each type should the merchant use in order to produce 30 kilograms of a blend that is worth $2.95 per kilogram?"

Even though this involves money we treat it like a mixture problem.

You got to unknowns but if you look at it as having x = 1 amount and y = 30 -x then you just have x, and 30 -x so...

Its the x amount * the 270 cent worth tea plus the 30 - x amount * the 300 cent worth tea is going to equal 30 * 295 cent blend

270 * x + 300(30-x) = 30(295)

270x + 9000 - 300x = 8850

-30x = -150

30x = 150

x = 150/30 = 5

so 5 kilograms of the 270 cent or $2.70 tea and 30 -5 or 25 kilograms of the 300 cent or $3.00 tea would need to be mixed together to make 30 kilograms of the $2.95 cent tea.

check 270 * 5 + 300 * 25 = 30*295

1350 + 7500 = 88508850 = 8850


http://algebra-word-problem-solutions.blogspot.com/search/label/Mixture%20Problem%20Solutions

http://algebra-word-problem-solutions.blogspot.com/2011/10/mixture-problem-7-solution.html#comment-form


http://www.addthis.com/bookmark.php?pub=klitschkofan&url=http://algebra-word-problem-solutions.blogspot.com/2011/10/mixture-problem-7-solution.html&title=Mixture%20Problem%207%20Solution

Tags: Mixture Problem Solutions, algebra word problem solutions



Thursday, March 13, 2008


This is sort of like the opposite of the last problem. Instead of diluting it your giving 100% pure alcohol to increase its level. As a fraction 100% is always just 1. anything else is always over a 100 so 30/100 = 3/10ths

So in this problem you would have:

(3/10)20 + 1(x) = (20 + x)(1/2)

6 + x = 10 + 1x/2 *multiply by 2

12 + 2x = 20 + x

x = 8

So 8 quarts of pure alcohol would need to be added to increase the % to 50% from 30%.

just to check:

(3/10)20 + 8 = (20 + 8) 1/2

6 + 8 = 14 is true










http://technorati.com/tag/algebra+word+problem+solutions

http://technorati.com/tag/Mixture+Problem+Solutions

http://www.addthis.com/bookmark.php?pub=klitschkofan&url=http://algebra-word-problem-solutions.blogspot.com/2009/10/mixture-problem-6-solution.html&title=Mixture%20Problem%20#%206%20Solution



75% is the same thing as 3/4 ths, which makes an easier setup than using .75 in the problem.

(3/4) * 8 gallons = 24/4 = 6 gallons--- so there is 6 gallons of pure alcohol in the 8 gallons.

The problem asks how much water needs to be added to dilute it to only 50% or so only half of the total quarts is alcohol.

A non-algebra way to think of it is to treat it like an average.

((3/4)*8 + 0(8) ) / (8 +x) = 1/2 which is basically saying if you have 8 total cups which is 75% alcohol how many extra cups (x) would you need to add to 8 to make 1/2 or 50%

Keep in mind that water is the same thing as saying 0% alcohol and then treat it like the other mixture problems.

So the problem would read:

(3/4)8 + 0(x) = (1/2)(8+x)

6 + 0 = 4 + 1/2(x) * multiply by 212 = 8 + x

x = 4

So 4 quarts of water would need to be added to dilute the alcohol solution to 50%.





This mixture problem deals with dollars instead of percentages. But, works the same way as before.

She has 10 pounds of $3 coffee for a total worth of $30 and x pounds of $5 coffee. The question asks of how much of the x $5 coffee is needed to mix with the 10 pounds of $3 coffee to make (10 + x) pounds of $4.50 coffee with a total value of $4.50 * (10 + x)







to set this up we would then have:

3(10) + 5x = (10 + x) * 4.50

30 + 5x = 45 + 4.50 x *multiply by 100 to get rid of decimal

3000 + 500x = 4500 + 450x

50x = 1500

x = 1500/50

x = 30

so 30 pounds of the $5 grade must be added to the 10 pounds of $3 grade to make $4.50 grade-- If you think about it that makes sense. If you had exactly the same amount of $5 coffee (10 pounds) as the $3 coffee-- then that would make an average grade of $4 coffee--- so you know it would have to be a lot more than just 10 pounds to increase the value close to $5-- which 30 pounds does.

But, just to check..

Does

30 + 5(30) = (10 + 30) * 4.50 ?

30 + 150 = 40(4.50)

180 = 180 correct





A non-algebraic approach would see that in order to have x amount of quarts of boric acid solution you could use averages to figure out how many quarts at 30% boric acid solution you would need to mix with 2 quarts of 10% boric acid solution.

When I say averages I mean if you scored 10% on 2 tests in school how many 30% tests would you need to get you a 20% average?






Well you might have seen that 20% is exactly half way between 10% and 30%. You already know that if you had just 1 test at 10% you would need 1 test at 30% to get an average of 20%---- But, you have 2 tests at 10% - so you need 2 test at 30% to get the 20.

This is analogous to the boric acid solution. You would need 2 quarts of 30% and mix it with the 2 quarts of 10% in order to get 20% boric acid solution (4 quarts total).

Algebraically you can set it up like you would an average..

From earlier you had (10% + 10%+ 30% + 30%)/4 = 20%

It should be pointed out theres a couple ways to work with percents. You can either write it in decimal: 10% becoming .10 , 50% becoming .5 or you can just write your percentage (minus the percent sign) over 100:10% becomes 10/100 , 50% becomes 50/100 for example

You can enter 10/100 and 50/100 into a calculator and you'll get the .10 and .5 respectively, which is the decimal version.

Once again from earlier we had (10% + 10%+ 30% + 30%)/4 = 20% which is the same thing as writing (10/100 + 10/100 + 30/100 + 30/100)/4 = 20/100 this equals...

(80/100)/4 = 20/100 =

(80/100) X (1/4) = 20/100 =

20/100 = 20/100 , which is correct since both sides of the "equals" are equal.

So to set this up with an unknown you have:

(10/100 + 10/100 + 30x/100)/(2+x) = 20/100

(20/100 + 30x/100) = (20/100) X (2+x) multiply by 100 and the 100's on the denominator cancel out

20 + 30x = 20 X (2+x) =20 + 30x = 40 + 20x =10x = 20 x = 20/10 = 2

So there you have an algebraic way of solving it...

As a shortcut you can skip right to the chase algebraically by setting up the problem as (# of quarts from Jar A) X ((Percentage of Solution) + (x amount of quarts of Jar B)) X (Percentage of Solution) = (# of quarts from Jar A + x amount of quarts from Jar B) X (Needed Percentage of

Solution)

in this case we had : 2 X (10/100) + x X (30/100) = (2+x) X (20/100)= 20/100 + 30x/100 = 40/100 + 20x/100 multiply by 100

* could have just gotten rid of the 100's from the beginning but you can make sure of no mistakes this way especially when doing other problems

For all other mixture problems you can see that the percentages when written over 100 allow you to cancel the 100's -- so for now on problems like this you can start wrting it as2(10) + 30x = 20(2+x) =

20 + 30x = 40 + 20x =10x = 20x = 2

OK I might have been long winded on this one, but I just wanted to say more than I had to on the 1st real mixture problem. I'll be less long winded on the next problem of this type.




A easy way to set up percentages is to always put that percentage as a number over 100. So 15% percent is the same thing as 15/100 ...

So if a mixture of 40 quarts has 15% alcohol then there is 40 X (15/100) = 4 X (15/10) = 60/10 =

6 quarts of alcohol in the mixture.

Number Problem 14 Solution

This is the solution for the Algebra number word problem which was asked by an anonymous visitor that read, "55 bowls of the same size capacity of food.1. everyone gets their own bowl of soup2. every two gets one bowl of spaghetti to share3. every three will get one bowl of salad 4. all r required to have their own helping of salad, spaghetti, and soup."

Ok you know the total number of people is x. Since '" every two gets one bowl of spaghetti to

http://algebra-word-problem-solutions.blogspot.com/2011/12/number-problem-14-solution.html






share" then x/2 of the people will have those bowls and since "every three will get one bowl of salad" then x/3 of the people will have those bowls in which they will total 55 bowls. So we have:

x + x/2 + x/3 = 55 ** multiply by 6

6x + 3x + 2x = 330

11x = 330

x = 30

So there are 30 bowls for soup, 30/2 = 15 bowls for spaghetti , and 30/3 = 10 bowls for salad.

Check 30 + 15 + 10 = 55


Labels: Number Problem Solutions



This is the solution to algebra number problem 13 as asked by an anonymous user: "A maths test contains 10 questions. Ten points are given for each correct answer and three points deducted for an incorrect answer. If Ralph scored 61, how many did he get correct?"

Ok on this problem you can come up with a solution faster by just quick trial an error. You know that if it only counts 3 points off for each wrong answer and they got a 61 then they had to have at least a 70 before the points were taken off. If they got a 70 that means they missed 3 problems (3 * 3 = 9) 70 - 9 = 61 . So the answer is they got 7 problems right and 3 wrong.

Now to set this up using algebra...

10x - 3(10-x) = 61 You are subtracting the wrong answers worth 3 points a piece from the right ones worth 10 a piece. You don't know how many of each so you are saying there are x amount of 10 valued answers and (10 - x) number of 3 valued wrong answers.

10x - (30 - 3x) = 61

10x -30 + 3x = 61

13x = 91 --> x = 91/13 = 7 --> so there are 7 right 10 valued problems and (10-7) or 3 wrong 3 point negative value ones.....


http://algebra-word-problem-solutions.blogspot.com/search/label/Number%20Problem%20Solutions

http://algebra-word-problem-solutions.blogspot.com/2011/12/number-problem-14-solution.html#comment-form


http://www.addthis.com/bookmark.php?pub=klitschkofan&url=http://algebra-word-problem-solutions.blogspot.com/2011/12/number-problem-14-solution.html&title=Number%20Problem%2014%20Solution

You could have also set it up like this 10(10 - x) - 3x = 61 and got the same answer....



Sunday, October 9, 2011


This is the solution to a number algebra word problem that was asked by Zakeus who said , "The numerator and denominator of a fraction are together equal to 100. Increase the numerator by 18 and decrease the denominator by 16 and the fraction is doubled. What is the fraction?"

Ok you want to set up 2 equations the 1st one is easy. x + y = 100 so y = 100 - x

The other would be (x + 18)/(y - 16) = (x/y)*2

(x + 18) y = 2x (y - 16) Now substitute 100-x for y....

(x + 18)(100 - x) = 2x(100 - x - 16)

100 x + 1800 - x^2 -18x = 2x(84-x)

82x + 1800 -x^2 = 168x - 2x^2 Now combine like terms...

x^2 - 86x + 1800 = 0

(x - 50)(x - 36) = 0 Now 50/50 is 1 and that's not the fraction we are looking for ( y = 100 - 50 = 50)

So substituting 36 in for x we have y = 100 - 36 = 64 So the fraction is 36/64 Now to verify this is correct add 18 to the numerator and subtract 16 from the denominator and we get ..

54/48 ... Ok now to verify if this is double the other equation lets reduce the equations. 36/64 4 is the gcd so we get 9/16 and for 54/48 , 3 is the gcd and we get 18/16 and 9/16 * 2 does equal 18/16, so we have the correct answer....














This is a new problem as asked by a anonymous visitor which read: "a number added to five times its reciprocal is -6. find the number "

Ok non-algebraically you can think of what numbers added together will make 6(of course they will be negative to make negative 6) --- 1 and 5 and 4 and 2 , and 6 and 0... You can't have a reciprocal of 0 and a reciprocal of 6 : 1/6 when multiplied by 5 makes 5/6ths and you know those added together : -6 + - 5/6ths dont make -6. Same thing goes for 2 and 3 --> 2 + 5/2 or 3 + 5/3 dont make 6. 1 and 5 however do work. The reciprocal of 1 is 1 so 1 + 5/1 =6 and so does 5 + 5/5 = 6 (Of course negative again). So there are 2 answers -1 and -5...

Algebraically... -x + (1/-x)5 = -6 =

-x -5/x = -6 (multiply by -x)

x^2 + 5 = 6x

x^2 + 6x + 5 = 0

(x + 1) (x + 5 ) = 0

So the answers are x = -1 and -5





This is number problem 10 solution which read:

"Find 3 consecutive even integers such that 4 times the first is decreased by the second is 12 more than twice the third. " Ok I'm not understading the "is decreased by the" part so I assume it was a typo and you just meant "Find 3 consecutive even integers such that 4 times the first is 12 more than twice the third. " ok 3 consecutive even integers would b x, x + 2, and x + 4

4 times the 1st would be just 4x and is equal to 12 more than twice the 3rd which would be (x + 4) * 2 + 12







which is... 4x = (x+4)*2 + 12

4x = 2x + 8 + 12

2x = 20

x = 10 So the answer would be 10, 12, 14 to check 4(10) = (14)*2 + 12

40 = 28 + 12

40 = 40





This is the solution to the algebra number word problem which was asked by a anonymous person in the ask question section which read, "One number is 5 more than another.their sum is 53.what are the two numbers?"

So we have one number is x and another is 5 more than it which is x + 5. they equal 53.

So we have x + x + 5 = 53

2x + 5 = 53

2x = 48

x = 48/2 = 24 for one number and 24 + 5 or 29 for the other.

So check 24 + 29 = 53

53 = 53











http://www.addthis.com/bookmark.php?pub=klitschkofan&url=http://algebra-word-problem-solutions.blogspot.com/2010/03/number-problem-9-solution.html&title=Number%20Problem%20%209%20Solution

Friday, March 5, 2010


This is the solution to the Algebra Number word problem # 8 which read "The ratio of 16 more than a number to 12 less than that number is 1 to 3. What is the number?"

Ok a number is x - the unknown

The problem says that ration of 16 more than a number --> which is x + 16 to 12 less than that number(the number is still x) or x - 12 is 1 to 3

So we have (x + 16)/(x -12) = 1/3Cross multiply and you get 3*(x + 16) = 1*(x - 12)

3x + 48 = x - 12

2x = - 60

x = -30

Check:(-30 + 16)/(-30 - 12) =-14/-42 = 14/ 42 = 1/3



Wednesday, October 7, 2009


This is the solution to the algebra number word problem 7 which asked "If the Numerator and Denominator of a certain fraction are both increased by 3, the resulting fraction equals 2/3. If the Numerator and Denominator are both decreased by 2, the resulting fraction equals 1/2. Determine the fraction."

Ok so you have 2 unknowns --> x the numerator and y the denominator. You can solve 2 unknowns with 2 equations. They give you 2 equations.

(x+3)/(y+3) = 2/3 and






http://www.addthis.com/bookmark.php?pub=klitschkofan&url=http://algebra-word-problem-solutions.blogspot.com/2010/03/number-problem-8-solution.html&title=Number%20Problem%20%208%20Solution

(x-2)/(y-2) = 1/2

take the (x+3)/(y+3) = 2/3 and cross multiply and get:

2(y+3) = 3(x+3) which is

2y + 6 = 3x + 9

Now take the second equation: (x-2)/(y-2) = 1/2 and cross multiply2(x-2) = 1(y-2) which makes

2x - 4 = y - 2 now solve for yy = 2x - 2

ok now plug this back into the 1st equation for y which we broke down to :2y + 6 = 3x + 9 so you get 2(2x-2) + 6 = 3x + 9 solve for x

4x - 4 + 6 = 3x+ 9x= 7 ok now plug 7 into y = 2x - 2 and get y = 2(7) -2 = 12

so the numerator is 7 and the denominator is 12 = 7/12

to check add 3 to the top and bottom and get 10/15 = 2/3

and subtract 2 from the top and bottom and get 5/10 = 1/2

Tags: Number Problem Solutions, algebra word problem solutions





This is the solution to the number problem # 6.

In the following Algebra number word problem we have a 3 digit number which has a 100's digit






http://technorati.com/tag/Number+Problem+Solutions


3 more than the 1's digit and its 10's digit is twice the 100's digit. When all 3 digits are added together, it is equal to 9. What is the number?

If you make the 1's digit x then you have:

100's digit = x + 3 and the

10's digit = 2(x + 3) or 2x + 6

Now when all these digits are added together you get 9 so...

x + x + 3 + 2x + 6 = 9

4x = 0

x = 0

so if x = 0(the 1's digit) then the 100's digit would be 0 +3 or just 3 and the 10's digit would be 0 + 6 or 6.

so the answer is 360

Technorati Tags: Number Problem Solutions, Algebra Word Problem Solutions



Monday, March 10, 2008


This one's a lot easier to work algebraically. You could get it by trial an error or course, but its easy if you think of one number being just x and the other number being (40-x)

so for example if the number was 10 then the other number would be (40 - 10) or 30

*this is pretty close if you check 5(10) = 50 2(30) = 60 which is a 10 difference and not a 3 through trial and error you could get the right answer just switching out a new number

Algebraically:





http://technorati.com/tag/Algebra+Word+Problem+Solutions

http://technorati.com/tag/Number+Problem+Solutions


So 2 times the 1st number lets use (40-x) since multiply by 2 with this is easier than by 5

2 (40 - x) is "equal" to

3 more than 5 times another number (x)

or 5x + 3

altogether we have 2(40-x) = 5x + 3

80 - 2x = 5x + 377 = 7xx = 11

so 1 number is 11 and the other number is (40 - 11) or 29

check it--- 2(29) = 58 5(11) = 55 and indeed they are 3 apart




You might be able to figure this out in your head since it doesn't matter if the number was 900, 9 million etc if its a situation that add up to 9 and 1 number is twice as big that you're adding. So what numbers 1 through the number 9 could you add together to get 9? 1 and 8 , 2 and 7, 3 and 6, and 4 and 5. Now which one of these is twice as big as the 1st number? The 6 is twice as big as the 3. But the problem asked for what 2 numbers added up to 900. So just add 2 zeros onto the 6 and 3. 300 and 600 are the answers.

Algebraically,

x = 1st number2x = 2nd number *2x means the same thing as 2 times x or 2 X x 900 is the number were looking for so..

x + 2x = 9003x = 900x = 900/3x = 300substituting back in x is 300 and 2x is 2(300) or 2 X 300 which is 600.











Odd numbers are 5, 11, 17, 19, 21, etc and consecutive means one right after the other as in 3,4,5,6, etc . So 3 consecutive odd numbers means 45,47,49 for example.

You want 3 consecutive odd that add up to 51 so the 1st number is x, the next number is x +2, and the last number is (x + 2) + 2 or x + 4. So,

x + x + 2 + x + 4 = 51 which equals

3x + 6 = 51 =

3x = 45

x = 45/3 = 15

So substituting back in for x you have 15 + (15 + 2) + (15 +4) = 51

15 + 17 + 19 does = 51 which is the answer ...



Saturday, March 8, 2008


Even numbers are 2, 6, 12, 18, etc- consecutive means one right after the other 1,2,3,4-- so consecutive even means 6,8,10,12, etc ...

So, the 1st number would be x, the 2nd number x + 2, and the 3rd number would be (x + 2) + 2 or x + 4 . The problem states when these 3 numbers are added together they equal 138.

We then have x + x + 2 + x + 4 = 138 which makes3x + 6 = 138 which makes3x = 132which makes x = 132/3 which is x = 44










Substituting back in we have 44 + 44 + 2 + 44 + 4 = 138 and that equals44 + 46 + 48 = 138 so 138 = 138

The 3 consecutive even numbers when added together that equal 138 are 44, 46, 48




Consecutive means 1 right after the other as in 3,4,5,6,7.... So in this problem the 1st number is x the next number is x + 1 and the last number is (x+1) +1 which is x + 2.

So we have x + x + 1 + x + 2 = 48 3x = 45 x = 15 ..... substituting back in the problem15 + (15 + 1) + (15 + 2) = 48 = 15 + 16 + 17 = 48 ... so the 3 consecutive numbers are 15,16,17.

Plane Geometric Figure Problem 3 Solution

This is the solution for the Algebra plane geometric figure problem as asked by an anonymous visitor which read, "catrina and tom want to buy a rug for a room that s 14 by 15 feet. They want to leave an even strip of flooring uncovered around the edges of the room. How wide a strip if they buy a rug with an area of 110 square feet?"

It's easier to picture this. Draw a rectangle that's almost a square 14 x 15 feet. Within in draw another such figure but smaller. The space between that figure and the 14 x 15 figure is x. Now the smaller square has an area of 110 square feet. We now have an equation :

(15 -x ) (14 -x ) = 110

210 + x^2 -14x - 15x = 110

x^2 - 29x + 210 = 110

x^2 -29x + 100 = 0

(x - 25)(x - 4) = 0

x = 4 or 25 however using 25 would result in negative dimensions which does not make sense. So 4 is the answer

http://algebra-word-problem-solutions.blogspot.com/2011/12/plane-geometric-figure-problem-3.html






Check: (15 - 4) (14 - 4) = 110 = 11*10 = 110


Labels: Plane Geometric Figure Problem Solutions



This is the Plane Geometric Figure Problem 2 Solution which asked "The length of the second side of a triangle is four less than three times the length of the first side. The length of the third side is one more than the length of the first side. If the perimeter of the triangle is 37 feet, what is the length of the first side?"

Since everything is relative to the 1st side make the 1st side x.

The 2nd side is 3x - 4 (4 less than 3 times the length of the 1st side)

and the 3rd side is x + 1 ( 1 more than the length of the 1st side)

Now all the sides add up to 37 feet so set all this equal to 37.

x + 3x - 4 + x + 1 = 37

5x - 3 = 37

5x = 40

x = 8

So the 1st side is 8 the 2nd side is 3(8) - 4 or 20 and the 3rd side is 9 .. 8 + 20 + 9 = 37


Labels: Plane Geometric Figure Problem Solutions

http://algebra-word-problem-solutions.blogspot.com/search/label/Plane%20Geometric%20Figure%20Problem%20Solutions

http://algebra-word-problem-solutions.blogspot.com/2010/11/plane-geometric-figure-problem-2.html#comment-form



http://algebra-word-problem-solutions.blogspot.com/search/label/Plane%20Geometric%20Figure%20Problem%20Solutions

http://algebra-word-problem-solutions.blogspot.com/2011/12/plane-geometric-figure-problem-3.html#comment-form


http://www.addthis.com/bookmark.php?pub=klitschkofan&url=http://algebra-word-problem-solutions.blogspot.com/2011/12/plane-geometric-figure-problem-3.html&title=Plane%20Geometric%20Figure%20Problem%203%20Solution

http://www.addthis.com/bookmark.php?pub=klitschkofan&url=http://algebra-word-problem-solutions.blogspot.com/2010/11/plane-geometric-figure-problem-2.html&title=Plane%20Geometric%20Figure%20Problem%202%20Solution

Sunday, August 2, 2009


This is the solution to the Plane Geometric Figure Problem 1 which asked you to "find the dimension of a rectangle whose length is 9 less than twice its width if the perimeter is 120 cm."

In a rectangle 2 times the width plus to times the length equals the total perimeter.

Let the width equal x

Then the length is equal to 2x - 9 which means "9 less than twice its width"

Remember 2 times the width plus to times the length equals the perimeter so..

2(x) + 2(2x - 9) = 120which is..

2x + 4x - 18 = 120

6x - 18 = 120

6x = 120 + 18

6x = 138

x = 138/6

x = 23

Since we let x stand for the width then the width of the rectangle is 23 cm and the length is 2(23) - 9 = 46 - 9 = 37

So the width is 23 cm and the length is 37 cm

Check 2(23) + 2(37) = 120

46 + 74 = 120120 = 120

This is the solution for a algebra time and distance word problem asked by a anonymous user: "Two trains left a station at the same time. One traveled north at a certain speed and the other traveled south at twice that speed. After 3 hours , the trains were 400 miles apart. How fast was each train traveling?"


The speed of the 1st train is x and the 2nd train is twice that or 2x. Ok the trick here is lets say someone is going 50 mph north and the other person is going twice that 100 the opposite direction. This distance separating them would be the same thing as if you added their speeds and it was going one direction -- in other words if someone was just going 150mph they would cover the same ground. Also rate multiplied by time equals distance. We know the time is 3 hours and the distance is 400. So

(x + 2x) (3) = 400

3x(3) = 400

9x = 400

x = 400/9 = 44 and 4/9ths mph is the 1st train and 88 and 8/9ths (800/9) mph is the 2nd train.


Labels: Time and Distance Problem Solutions

Sunday, January 10, 2010

Time and Distance Problem 11 Solution

This is the answer to the algebra time and distance word problem #11 which was asked by another anonymous user on the "Algebra Word Problem Questions" post. The question read:

"Alexis left Miami and drove at a speed of 20 kph. Thomas left 3 hours later, from the same point, and drove at a speed of 30 kph. How long will it take Thomas to catch up to Alexis?"

Ok non-algebraically the easiest way to figure this one out is to figure how far kilometers Alexis has gone before Thomas even leaves. Since Thomas left 3 hours later and Alexis drove 20 kph then Alexis has gone 20(3) or 60 k or kilometers. Thomas is going faster but how much faster than Alexis? (30 - 20)kmh = 10 So, the question is if he's creeping up at 10 kmh how long would it take him to go the distance that Alexis is ahead which we figured out to be 60 kilometers. Well at 10 kph it would take 6 hours to to 60 k... 60 k / 10 kph = 6 h

For the algebra way you just set up the equation where you have :

20(3 + x) = 30x --> the time is x + 3 because she had a 3 hour head start

then you have 60 + 20x = 30x

60 = `10x

http://algebra-word-problems.blogspot.com/2009/08/algebra-word-problem-questions.html

http://algebra-word-problem-solutions.blogspot.com/2010/01/time-and-distance-problem-11-solution.html

http://algebra-word-problem-solutions.blogspot.com/search/label/Time%20and%20Distance%20Problem%20Solutions

http://algebra-word-problem-solutions.blogspot.com/2011/10/time-and-distance-problem-13-solution.html#comment-form


http://www.addthis.com/bookmark.php?pub=klitschkofan&url=http://algebra-word-problem-solutions.blogspot.com/2011/10/time-and-distance-problem-13-solution.html&title=Time%20and%20Distance%20Problem%2013%20Solution

x = 60/10 = 6 --> 6 hours

Tags: Time and Distance Problem Solutions, algebra word problem solutions


Labels: Time and Distance Problem Solutions

Saturday, September 12, 2009

Time and Distance Problem 10 Solution

This is the solution to the algebra time and distance word problem # 10 which was submitted under the ask a question post anonymously which read..

"Junior's boat will go 15 miles per hour in still water. If he can go 12 miles downstream in the same amount of time as it takes to go 9 miles upstream then what is the speed of the current?"

distance = rate X time ... d = r*t we know the speed of the boat in still water 15 mph

we know he can go 12 miles downstream in the same time he can go 9 miles upstream-- so times are equal --- so since time is equal we can set the distances and rates equal to one another...

d1/r1 = d2/r2

The rate downstream will be 15 + x and upstream will be 15 - x so we have:

12/(15+x) = 9 /(15-x)

cross multiply and you get 12(15-x) = 9 (15+x)

180 - 12x = 135 + 9x

45 = 21x

x = 45/21 = 15/7 This is the answer the current is 15/7 or 2 and 1/7ths mph..


http://algebra-word-problem-solutions.blogspot.com/search/label/Time%20and%20Distance%20Problem%20Solutions

http://algebra-word-problem-solutions.blogspot.com/2010/01/time-and-distance-problem-11-solution.html#comment-form



http://technorati.com/tag/Time+and+Distance+Problem+Solutions

http://www.addthis.com/bookmark.php?pub=klitschkofan&url=http://algebra-word-problem-solutions.blogspot.com/2010/01/time-and-distance-problem-11-solution.html&title=Time%20and%20Distance%20Problem%2011%20Solution


Work Problem 6 Solution

This is the solution to Work Problem 6 which asked, "A mother can rake a yard in 90 minutes and her daughter can do it in 60 minutes.If the mother rakes for 15 minutes before her daughter joins her,how long will it take them to finish the work?"

Ok the mother can do 1/90th of her job in a minute so if you have x/90 in 90 minutes 90/90 = 1 she completes her job same thing goes for the the daughter .. 1/60 of her job done in a minute with x/60 in 60 minutes 60/60 = 1 she completes her job. Seems kind of silly that I'm pointing this out but this is how you solve the problem you just add their work together and set it equal to 1. Their is one additional stipulation in the problem though and thats the mother starting 15 minutes before the daughter joins the mother otherwise it would just bex/90 + x/60 = 1... but the mother rakes for 15 minutes and completes 15/90 of her job so the way to write it out would be x/90 + 15/90 + x/60 = 1.

multiply everything by 180 and get 2x + 30 + 3x = 180.

5x + 30 = 1805x = 150x = 30 So it would take 30 minutes if they worked together. UPDATE: "how long will it take them to finish the work?" <-- Sounds to me like how long will they finish together thus 30 minutes the answer. But if you want to take this to mean how long did it take them to finish the total job then add the 15 minutes of initial work done by the mother which would be 30 + 15 = 45.


Labels: Work Problem Solutions

Tuesday, August 17, 2010


This is the solution to the work problem # 5 which read: Jim can fill a pool carrying buckets of water in 30 minutes. Sue can do the same job in 45 minutes. Tony can do the same job in 1 ½ hours. How quickly can all three fill the pool together?

Ok Tony takes 1 and 1/2 hours which is 90 minutes, and then we got jim at 30 minutes and Sue at 45 minutes. Convert everything to find out how much each can do in 1 minute. Tony does 1/90 th of his job in 1 minute Jim 1/30 th and Sue 1/45 th. So lets take just one individual persons job say Sue for example. If someone says Sue takes how long to finish her job if you know that she

http://algebra-word-problem-solutions.blogspot.com/2010/08/work-problem-5-solution.html

http://algebra-word-problem-solutions.blogspot.com/search/label/Work%20Problem%20Solutions

http://algebra-word-problem-solutions.blogspot.com/2010/11/work-problem-6-solution.html#comment-form



http://www.addthis.com/bookmark.php?pub=klitschkofan&url=http://algebra-word-problem-solutions.blogspot.com/2010/11/work-problem-6-solution.html&title=Work%20Problem%206%20Solution

can do 1/45 of her job in a minute what would u say? 45 minutes. Or you can look at it as x/45 = 1 which means how many over 45 would = 1? The answer of course would be 45 again. So we set this problem up just like that only were ADDING the other peoples jobs together and setting it equal to 1.

So we got x/45 + x/90 + x/30 = 1 *Multiply both sides by 90 and get

2x + x + 3x = 90

6x = 90

x= 90/6 = 15

So it would take 15 minutes if all 3 worked together to finish the job.



Saturday, November 14, 2009


This is the solution to the algebra work word problem # 4 which read: "Joe and Jim paint a house. Joe can paint it alone in 5 days, Jim in 8 days. They start to paint it together but after 2 days Jim stops. How long will Joe finish it alone?"

Joe takes 5 daysJim 8 to paint a house

So Joe can do 1/5 of his job in one day and Jim can do 1/8th of his job in 1 day. If you just take Joe for example you can see where I'm heading on solving this problem. Joe can do 1/5th of his job in one day. So how long would he finish his job -- well besides knowing the answer already -- you would set it up as x/5 = 1. 1 being when the job is complete not just a "fraction" of his work being done. So the answer is 5 of course 5/5 = 1. You use this exact same method to solve the problem for both Joe and Jim.

**********Correction************ I realized I solved the problem before if they both continued working the job, but this problem states JIM stops working in 2 days. So Joe does x/5 part of his job in 2 days? 2/5 and Jim does x/8 part of his job in 2 days? 2/8, and Joe keeps working. So the equation would be

2/5 + 2/8 + x/5 = 1 multiply by 40 and get16 + 10 + 8x = 40



http://algebra-word-problem-solutions.blogspot.com/2010/08/work-problem-5-solution.html#comment-form



26 + 8x = 408x = 40 - 268x = 14x = 14/8 or 1 and 3/4ths so it would take 1 and 3/4ths days more for Joe to finish the job alone..

Tags: Work Problem Solutions, algebra word problem solutions


Thursday, February 5, 2009


In this Algebra work problem solution we have a similar situation in dealing with hours and and figuring out how much of a job is done in 1 hour. The only difference is part of the job is being taken away -- negative --- by the leak. So just subtract this. That's the only difference between this work problem and the others.

So with Pump A in 1 hour it would 1/12th of a job Pump B would do 1/6th of a job and the leak would take away 1/24th of the job.

So algebraically we have:

x/12 + x/6 - x/24 = 1

multiply by 24

2x + 4x - x = 24

5x = 24

x = 24/5 or 4 and 4/5ths

4/5ths of an hour is 60*4/5 or 240/5 = 48 minutes

So the pool would be filled in 4 hours and 48 minutes.




http://technorati.com/tag/Work+Problem+Solutions



Sunday, February 1, 2009


This is the solution to the Algebra work problem # 2.

The key to solving any work problem is to figure out how much work can be done in 1 hour, minute, day- whatever time classification your using. In this problem it's talking hours. Since pump A takes 10 hours to finish then in 1 hours times 1/10th of the work would be complete and with the 8 hour pump B, 1/8th of the work would be done.

Now the way to understand this is to look at just 1 of the pumps. Take pump A for example. It takes 10 hours to fill the pool and 1/10 of the pool is finished in 1 hour. So ask yourself how many hours would it take a pump that fills 1/10th of a pool an hour to fill the entire pool? Or X/10 = 1 --- Without Algebra you know its 10 but using algebra you can see it works out the same way.

So all we do when were dealing with more than just 1 pump is to add their respective ratios together and set them equal to 1. We add because it's the added effect of both pumps that fills the pool.

So we have algebraically:

X/10 + X/8 = 1

multiply by 40 ----- the common denominator....

40X/10 + 40X/8 = 40

4X + 5X = 40

9X = 40

X = 40/9

X = 40/9 or 4 and 4/9ths

so almost 4 and a half hours. To get the exact minutes take 60 * 4/9 = 240/9 = 26 and 2/3rds

so we got 4 hours 26 minutes and if u want to be real anal 2/3's of a minute is 40 seconds (60 * 2/3 = 120/3 = 40)

So, it takes 4 hours 26 minutes and 40 seconds for pumps A and B to fill up the pool if they work



together.


Sunday, September 28, 2008


In this Algebra work word problem solution the way to figure out the answer is to realize that they work a fractional part of work each hour. So in 1 hours time Steve does 1/9 of his work, so in 9 hours he does 9/9 or his whole job. Same thing goes for Joe except he gets 1/10th of his job done in an hour. For just one guy, take Steve for instance, the algebra for his work by himself would be (1/9)x = 1 or x/9 = 1 which makes x = 9 or 9 hours. for Joe it would be x/10 = 1 or 10 hours. Together they would take x/9 + x/10 = 1 hours(x) to get the job done.

x/9 + x/10 = 1 multiply by 9010x + 9x + 9019x = 90x = 90/19

90/19 = 4.74 or approximately 4 hours and 45 minutes (.75 would 3/4ths of an hour or 45 minutes). If you want it exact just take the fractional part(.7368) and multiply by 60. You'll end up with 44.2 which is 44 minutes and .2*60 seconds or 12 seconds. The detailed answer would be 4 hours 44 minutes and 12 seconds. So if he hires both of these guys he'll get the lawn mowed before 6 hours.

