Solving a Tricky Math Question: Peter and Lisa's Ages

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In summary, Peter is 4 years old, Lisa is 7 years old, and their father is 28 years old. The equations to solve this problem are: Peter + Lisa = 11, Peter * Lisa = Father's age, and Father's age = Lisa * 8. After solving, it is determined that Peter is 4 years old, Lisa is 7 years old, and their father is 28 years old.
  • #1
amix
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Hey guys.

I have this tricky math question.

Peter and his big sister Lisa are together 11 years old. The product of their age is equal to the age of their father. When Peter was born the father was 8 times older than Lisa. How old are Lisa and Peter?

You can put up the info in the text like this:
1. Peter + Lisa = 11
2. Peter * Lisa = Fathers age
3. Fathers age = Lisa * 8

Well I have tried to solve this but without any help.

Anybody know how to do this?

Thanks.
 
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  • #2
You've made a good start but the last equation is wrong.


The last sentence does not say that the father's age NOW is 8 times is 8 times Lisa's age- and that is what you are using "father's age", "Peter" and "Lisa" to represent in the first two equations.

If "Peter" is used to represent Peter's age now, then Peter was born "Peter" years ago (Yes, that looks silly- that's one reason why it's often better to use letters rather than words!).

The father's age when Peter was born was:
"father's age- Peter". At that time Lisa's age was "Lisa- Peter".
The last equation is "father- Peter= 8*(Lisa- Peter)"

Here's how I would do the problem myself:

Let P= Peter's age now, L= Lisa's age now, F= father's age now.

Then the first sentence, "Peter and his big sister Lisa are together 11 years old.", becomes the equation "P+ L= 11".

The second sentence, "The product of their age is equal to the age of their father.", becomes "L*P= F".

These two equations are exactly the same as your first two equations.

The third sentence, "When Peter was born the father was 8 times older than Lisa.", becomes "F- P= 8*(L- P)".
That last equation is equivalent to F- P= 8L- 8P which is also equivalent to F= 8L- 7P.

The first equation, P+ L= 11 is the same as L= 11- P (Subtract P from both sides).

If we replace the L in F= 8L- 7P by 11-P, we get F= 8(11-P)-7P
= 88- 8P- 7P= 88- 15P: that is, F= 88- 15P.

Now, replace F and L in L*P= F by those:

L*P= (11-P)*P= 88- 15P which is the same as
11P- P<sup>2</sup>= 88- 15P

adding P<sup>2</sup> and subtracting 11P from both sides gives

P<sup>2</sup>- 26P+ 88= 0.

That's a quadratic equation for P. You can solve that by using the quadratic equation, completing the square, or factoring (factoring is simplest IF the polynomial factors- here it DOES!).

Once you have found Peter's age (P), you can use L= 11- P to find Lisa's age and F= 88- 15P to find the father's age.

Be sure to check that your answers make the original sentences true!
 
  • #3
Ok, I didn't understand the previous solution, so I'll post my own... but I've always been horrible at maths...

Peter's age - x.
Lisa's age - y.
Father's age - z.

x + y = 11
xy = z
8 (y - x) + x = z

xy = 8y - 7x
11x - x^2 (squared) = 88 - 8x - 7x
x^2 - 26x + 88 = 0
x1 = 22 (doesn't fit)
x2 = 4

y = 11 - 4 = 7
4 * 7 = z = 28

Peter's age = x = 4
Lisa's age = y = 7
Father's age = z = 28
 
Last edited:
  • #4
BTW

(y - x) is the difference in age between lisa and peter
 
  • #5
thanks allot guys.
I really could use you help, and I appreciate it!
 
  • #6
Tail
Umm ... weird.
Your third equation is wrong, it should be :
8(y-x)=(z-x)
BUT[/B} the first line in your solution (which is supposed to be based on the second and third equations) is right, so i suppose you just made a mistake while writing the third equation.
Otherwise, your answers are right :smile:.
 
  • #7
I edited the post... a little mistake, as I did the calculations on paper and then typed them...
 
  • #8
Staii--- Have you been staying up too late?

Tail's "third equation" 8(y- x)+ x= z is exactly equivalent to
8(y-x)= z- x (and to my "F- P= 8*(L- P)").

True, a direct translation of the sentence gives 8(y-x)= z- x but
Tail just went ahead and added x to both sides.
 
  • #9
amix said:
Hey guys.

I have this tricky math question.

Peter and his big sister Lisa are together 11 years old. The product of their age is equal to the age of their father. When Peter was born the father was 8 times older than Lisa. How old are Lisa and Peter?

You can put up the info in the text like this:
1. Peter + Lisa = 11
2. Peter * Lisa = Fathers age
3. Fathers age = Lisa * 8

Well I have tried to solve this but without any help.

Anybody know how to do this?

Thanks.

PETER is 4
LISA is 7
Their father was 24 when Peter was born, but at that time Lisa was 3.
Then 4 years later Lisa is 7 and Peter is 4.
Now since it is 4 years later their father is 28
7 times 4 is equal to 28.
 

1. How can I solve a tricky math question involving multiple variables?

The key to solving a tricky math question is to break it down into smaller, more manageable parts. Start by identifying all the given information and variables, and then use equations and logic to solve for each variable one at a time.

2. What are the steps for solving a math problem involving ages?

The first step is to clearly define the given information and the unknown variables. Then, create equations based on the relationships between the variables. Next, solve for the unknown variables using algebra or other mathematical methods. Finally, check your solution by plugging in the values and ensuring that the equations are balanced.

3. How can I determine the relationship between two people's ages in a math problem?

In most cases, the relationship between two people's ages can be represented by a simple equation. For example, in the problem of Peter and Lisa's ages, it can be represented as Peter's age + Lisa's age = total combined age. You can also use logic to determine the relationship, such as if one person is twice as old as the other.

4. What should I do if I get stuck on a tricky math question?

If you get stuck on a tricky math question, take a step back and try to approach it from a different angle. You can also break down the question into smaller parts or try to solve for a different variable first. If you still can't solve it, don't be afraid to ask for help from a teacher, tutor, or classmate.

5. Are there any common mistakes to avoid when solving a tricky math question?

One common mistake is not carefully defining the given information and variables, which can lead to incorrect equations and solutions. Another mistake is not checking your solution for accuracy. It's also important to keep track of units and make sure they are consistent throughout the problem.

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