8TH GRADE MATH HELP - SOLVING WORD PROBLEMS MADE EASY
Please study
Linear Equations in Two Variables before 8th Grade Math Help
if you have not already done so.
There, we explained the method of solving the equations, with examples.
That knowledge is a prerequisite here.
Here, we apply that knowledge to solve word problems.
Example 1 of 8th Grade Math Help
Solve the following Word Problem on 8th Grade Math Help.
A man travels 600km partly by train and partly by car.If he
covers 120km by train and the rest by car, it takes him 8 hours.
But, if he travels 200km by train and the rest by car, he takes
20 minutes longer. Find the speed of the car and that of the train.
Solution to Example 1 of 8th Grade Math Help :
Let x km/hr and y km/hr be the speeds of the train and car respectively.
We know
time = distance⁄speed
If the man covers 120km by train and
the rest (= 600 - 120 = 480km) by car,
then the time taken for total journey
= time taken for travelling 120km by train
+ time taken for travelling 480km by car
= 8 hours. (by data)
time taken for travelling 120km by train = distance⁄speed
= 120km⁄xkm/hr = 120⁄x hours.
time taken for travelling 480km by car = distance⁄speed
= 480km⁄ykm/hr = 480⁄y hours.
∴ 120⁄x + 480⁄y = 8
Dividing throughout by 8, we get
15⁄x + 60⁄y = 1............................(i)
If the man covers 200km by train and
the rest (= 600 - 200 = 400km) by car,
then the time taken for total journey
= time taken for travelling 200km by train
+ time taken for travelling 400km by car
= 8 hours + 20 minutes (by data) = 8 hours + (20⁄60)hours
= (8 + 1⁄3) hours. = 25⁄3 hours.
time taken for travelling 200km by train = distance⁄speed
= 200km⁄xkm/hr = 200⁄x hours.
time taken for travelling 400km by car = distance⁄speed
= 400km⁄ykm/hr = 400⁄y hours.
∴ 200⁄x + 400⁄y = 25⁄3
Dividing throughout by 25, we get
8⁄x + 16⁄y = 1⁄3
Multiplying throughout by 3, we get
24⁄x + 48⁄y = 1.............................(ii)
Equations (i) and (ii) are the Linear Equations
in two variables formed by converting the given
word statements to the symbolic language.
Now we have to solve these simultaneous equations.
let us denote 1⁄x by p and 1⁄y by q.
Then equations (i) and (ii) become
15p + 60q = 1.........................................(iii)
24p + 48q = 1.........................................(iv)
(iii) x 4 gives 60p + 240q = 4..........................................(v)
(iv) x 5 gives 120p + 240q = 5.........................................(vi)
Subtracting (vi) from (v), we get
60p - 120p = 4 - 5 ⇒ -60p = -1 ⇒ p = 1⁄60.
Using this value of p in (v), we get
60(1⁄60) + 240q = 4 ⇒ 1 + 240q = 4 ⇒ 240q = 4 - 1
⇒ 240q = 3 ⇒ q = 3⁄240 = 1⁄80
But p and q stand for 1⁄x and 1⁄y respectively
∴ p = 1⁄x = 1⁄60 ⇒ x = 60
and q = 1⁄y = 1⁄80 ⇒ y = 80
∴ speed of the train = x = 60km/hr. Ans.
and speed of the car = y = 80km/hr. Ans.
Check:
Time taken for 120km by train and the rest ( = 480km) by car
= 120km⁄60km/hr + 480km⁄80km/hr = 2hr. + 6hr.
= 8 hours. which is same as given in the problem. (verified.)
Time taken for 200km by train and the rest ( = 400km) by car
= 200km⁄60km/hr + 400km⁄80km/hr = (10⁄3)hr. + 5hr.
= {(3 + 1⁄3) + 5} hr. = 8 hr. + (1⁄3) hr. = 8 hr. + (1⁄3) x 60 minutes.
= 8 hr. + 20 minutes. which is same as given in the problem. (verified.)
Thus the Solution to Example 1 of
8th Grade Math Help is verified.
Example 2 of 8th Grade Math Help
Solve the following Word Problem on 8th Grade Math Help.
A motor boat takes 6 hours to cover 100 km downstream and 30 km
upstream. If the motor boat goes 75 km downstream and returns
back to its starting point in 8 hours, find the speed of the motor boat
in still water and the rate of the stream.
Solution to Example 2 of 8th Grade Math Help :
Let x km/hr. be the speed of the motor boat in still water.
and y km/hr. be the rate of flow of the stream.
Speed upstream = (x - y) km/hr.
Speed downstream = (x + y) km/hr.
We know, Time = Distance⁄Speed.
Time for 100 km downstream and 30 km upstream
= 100 km⁄(x + y) km/hr + 30 km⁄(x - y) km/hr
By data this = 6 hours.
∴ 100⁄(x + y) + 30⁄(x - y) = 6
Let us denote 1⁄(x + y) by p and 1⁄(x - y) by q.
The equation becomes 100p + 30q = 6
⇒ 50p + 15q = 3..................................(i)
Time for 75 km downstream and returning (means 75 km upstream)
= 75 km⁄(x + y) km/hr + 75 km⁄(x - y) km/hr
By data this = 8 hours.
∴ 75⁄(x + y) + 75⁄(x - y) = 8
Since we denoted 1⁄(x + y) by p and 1⁄(x - y) by q,
The equation becomes 75p + 75q = 8
⇒ 75p + 75q = 8..................................(ii)
Equations (i) and (ii) are the Linear Equations
in two variables formed by converting the given
word statements to the symbolic language.
Now we have to solve these simultaneous equations.
This Solving is an important topic in 8th Grade Math Help.
To solve (i) and (ii), let us make the coefficients of q the same.
(i) x 5 gives 250p + 75q = 15.....................................(iii)
(iii) - (ii) gives 250p - 75p = 15 - 8 ⇒ 175p = 7
⇒ p = 7⁄175 = 1⁄25
Using this in (i), we get
⇒ 50(1⁄25) + 15q = 3 ⇒ 2 + 15q = 3 ⇒ 15q = 3 - 2
⇒ 15q = 1 ⇒ q = 1⁄15.
But p = 1⁄(x + y) = 1⁄25 ⇒ x + y = 25..............(iv)
and q = 1⁄(x - y) = 1⁄15 ⇒ x - y = 15..............(v)
(iv) + (v) gives x + x = 25 + 15 ⇒ 2x = 40 ⇒ x = 20
Using this in (i), we get
20 + y = 25 ⇒ y = 25 - 20 = 5.
Thus, the speed of the motor boat in still water = 20 km/hr. Ans.
and rate of flow of the stream = 5 km/hr. Ans.
Check:
Time for 100 km downstream and 30 km upstream
= 100⁄(20 + 5) + 30⁄(20 - 5) = 100⁄25 + 30⁄15
= 4 + 2 = 6 hours which is same as per data. (verified.)
Time for 75 km downstream and returning (means 75 km upstream)
= 75⁄(20 + 5) + 75⁄(20 - 5) = 75⁄25 + 75⁄15
= 3 + 5 = 8 hours which is same as per data. (verified.)
Thus the Solution to Example 2 of
8th Grade Math Help is verified.
Exercise on 8th Grade Math Help
Solve the following Word Problems on 8th Grade Math Help.
- A train covered a certain distance at a uniform speed.
If the train had been 6 kmph faster, it would have taken
4 hours less than the scheduled time. And, if the train
had been slower by 6 kmph, it would have taken 6 hours
more than the scheduled time.
Find the length of the journey.
- A sailor goes 8 km downstream in 40 minutes and returns
back to the tarting point in 1 hour. Find the speed of the
sailor in still water and the speed of the current.
Perform the verification check on the
above problems on 8th Grade Math Help.
Answers to Exercise on 8th Grade Math Help
The Answers to the Word Problems on
8th Grade Math Help given
in the exercise above, are given below.
- 720 km.
- 10 kmph, 2kmph.
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