How to Solve Math Word Problems

A lot of students struggle with solving Word Problems. The objective of this lesson is to help you understand what exactly the word problem is asking and how to go write that in Mathematical terms so that you can solve it.

Remember that no matter how hard a word problem appears, the simple fact is that there will always be words to help you solve it!!! You already know those words,all that remains is for you to write it down correctly.

What they Say/What they Mean
The people who set Word problems have only 1 goal–To confuse you. So they try to use words that appear confusing.Unfortunately, they are limited by the rules of Mathematics. The rules are clear and simple. There is/can be no confusion about them.

So now we know what certain terms in a problem mean, let’s take a look at solving some of these problems.

How to Solve Age Problems:

Q.Four years from now,Peter’s age will be 4 times his son’s age. Twelve years from now he will be 2.5times his son’s age.Find their present ages.

Solution:
“Hmm, I’ve read the problem but I dont know where to begin”. We’ve heard that far too many times. When you dont know something, like the Detectives in CSI/Criminal Minds/(Insert favourite Detective Serial here) just put down X. If there’s more, use another alphabet say Y!
So,
Let Peter’s Present Age = X
His Son’s Present Age = Y

Four years from now
Peter’s age = X+4
His son’s age = Y+4

Now go back to the problem and read the first line

“Four years from now,Peter’s age will be 4 times his son’s age”

“Four years from now,Peter’s age” +”will be” +”4 times his son’s age”
_____________”X+4″___________”=”________ “4(Y+4)”
(Remember that the first line said 4 years from now which means that even the son would have aged by 4 years)

X+4 = 4(Y+4)
=>X-4Y = 12 —(1)

“Twelve years from now he” +”will be” + “2.5times his son’s age” __________”X+12″________”=”_______2.5(Y+12)

X+12=2.5(Y+12)=>2X-5Y = 36 —(2)

We know have 2 equations and 2 unknowns.Which is good. For all the unknows in a problem, we require that many equations. So if there were 3 unknowns,we would require 3 equations to solve them.

Solving (1) and (2) we get
y =4 which means X = 28

Q2.The present age of a father is 10times his son’s age. In 6 years, his age will be four times his son’s age. What are their present ages?

Solution:

As always, we start by defining the unknown
Let the father’s present age = X
Son’s Present age = Y

“The present age of a father is 10times his son’s age”

Whenever you get stuck converting something like “age is 10times”, then just use numbers, eg

If the son was 2, then 10times his age would be (2+2+2…10times OR 2*10) –> 20
so, IF he was y, then 10times his age would be (y+y….10times OR 10*y) –> 10y
X =10Y

“In 6 years, his age” +”will be”+ “four times his son’s age”
_______”X+6″______=________4 (Y+6)
X+6=4(Y+6)
=>X-4Y = 18

We now have our required two equations
X =10Y
X-4Y = 18

Solving we get
X = 30 and Y = 3

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How to Solve Math Clock Problems

Even a stopped clock is right twice a day.

-Anonymous  
That’s no help to if you are late to school is it??

To be able to solve problems about clocks quickly you must remember certain points. We’ve summarised those points below. Most are self-explanatory and for the rest, please accept our word and then work out on your own the how and the why.

The face(or dial) of a clock is divided along its circumference into 60 equal spaces called minute spaces. The minute hand moves around the whole circumference of the clock once in One Hour. The hour hand moves around the whole circumference of the clock once in 12 hours.
Thus, the minute hand is 12times faster than the hour hand.
The minute hand passes over the 60minutes spaces, while the hour hand goes over the 5 minute spaces.

That is, in 60minutes, the minute hand gains 55minutes over the hour hand or 55/60 minute spaces in 1 minute.

Some Important Points

• 1 minute space = 360/60(Since 360degrees of the circle is divided into 60minutes.
• In 1 minute, the minute hand moves 6degrees
• In  one minute, the hour hand moves 360/(12 x 60) = 360/720 = 0.5degrees

In 1 minute the minute hand gains 5.5degress over the Hour hand

• In ever hour,

1. The hands Coincide once  i.e. they are 0 degrees apart
2. The hands are twice at Right Angles (90degrees apart) .In these positions, the hands are 15 minute spaces apart.
3. The hands point in Opposite direction(180degrees apart)once.In this position , the hands are 30minute spaces apart.

• The hands coincide 11 times in every 12 hours ( between 11 and 1 O’clock there is a common position at 12). Hence in a day the hands coincide 22 times
• The hands coincide every 65(5/11 ) minutes

To Find the Angle Between the Hands

Angle Theta = |30h - (11m/2)|    where h = hour and m =minutes

Q.At what time between 4 and 5 will the hands of a watch coincide?

Solution:
Applying the above formula,  we know that h has to equal 4 since between 4 and 5 the time will obviously read 4:__
Now if the hands coincide it means that the angle between them is 0.

So we finally have 0 = |30(4) -(11m/2)|

120 = 11m/2
or m = 240/11 minutes  = 21 (9/11)minutes
which means at 21 (9/11)minutes past 4, the hands will coincide

Can you do solve for when the hands will be at Right Angles and when the hands are in Opposite Direction?

Incorrect Clocks:

A clock which gains or loses time is called an Incorrect clock.In incorrect clocks, both hands coincide at an interval not equal to 65(5/11) minutes.

For Slow Clocks(for clocks that lose time)

Total time lost in T hours = (T x 60) { X-65(5/11)}/X minutes
where X is the time in which the hands of the incorrect clock coincide.

For Fast Clocks(for clocks that gain time)

Total time lost in T hours = (T x 60) { 65(5/11)-X}/X minutes
where X is the time in which the hands of the incorrect clock coincide.

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