Lesson: Writing Equations

Comment on Writing Equations

Brent,

Brent,
The question and the narrative are both different in the example at time 3:00. The last paragraph is stated as follows: If Zoe is 3 times as old as Liam, how old is Liam. Based on the original derivation. G = Gita's age, 2G + 5 = Liam Age, then 3(2G + 5) = Zoe. So, How come we have 10G - 13 as Zoe's age. Please, explain.

Great question!

Great question!

Notice the question tells us TWO things about Zoe's age:

1) Zoe's age is 13 years less than 10 times Gita's age.
2) Zoe is three times as old as Liam.

These two pieces of information allow us to create an EQUATION.

From 1) we can write: Zoe's age = 10G - 13
From 2) we can write: Zoe's age = 3(2G + 5)

From these two facts, we can create the equation: 10G - 13 = 3(2G + 5)

Hi Brent for your explanation to the question on GMAT Club:
Company Q plans to make a new product next year and sell each unit of this new product at a selling price of \$2. The variable costs per unit in each production run are estimated to be 40% of the selling price, and the fixed costs for each production run are estimated to be \$5,040. Based on these estimated costs, how many units of the new product will Company Q need to make and sell in order for their revenue to equal their total costs for each production run?

A. 4,200
B. 3,150
C. 2,520
D. 2,100
E. 1,800

Instead of a decimal we can keep the total selling price as a fraction so that the task becomes much easier and there's no need for estimation (according to me)
Then the equation becomes:
2x=5040+4x/5
to get 6x=25200
therefore, x=4200 (Ans. A)
Is that correct? Thanks

Your approach is perfectly valid - nice work.

Hi Brent,

Hi Brent,
I tried substituting the number of bananas to get A=9- 5B/7 and substituted this in the main equation 7A+5B=63 but the calculation just gets more and more difficult after this.

Quote: I tried substituting

Quote: I tried substituting the number of bananas to get A = 9 - 5B/7 and substituted this in the main equation 7A + 5B = 63

The equations A = 9 - 5B/7 and 7A + 5B = 63 are EQUIVALENT EQUATIONS, so substituting one into the other isn't going to help.

I should mention that we can't really solve this question using regular algebra.

In high school we learned that, if we're given 1 equation with 2 variables, we cannot find the value of either variable. However, if we restrict the variables to POSITIVE INTEGERS, then there are times when we can find the value of a variable if we're given 1 equation with 2 variables.

ASIDE: You can see my full solution at http://www.beatthegmat.com/algebra-t297095.html

Thanks Brent! That made it much easier to understand

Hey Brent,

Hey Brent,
Sorry for bombarding you with so many questions but I really find your inputs really helpful.

For question https://gmatclub.com/forum/class-b-has-50-more-students-than-class-a-number-of-girls-in-class-a-242136.html
I found a very simple ultra easy way of solving this, so easy in fact that I almost find myself doubting the correctness. Please verify cause it practically ignores some info in the question stem.

Let there be 100 Students in Class A
Therefore 150 students in Class B

Let 'x' be the no of girls in class A = no of boys in Class B.

Therefore, the number of Boys in Class A = 100-x.
And the Boys in Class B = x

Adding the two together we get Total no of Boys
= 100-x + x= 100.

Total No of students = 150+100= 250.

Therefore, percent of boys in both the classes =
100 x 100 / 250 = 40%

Thanks, please let me know if there is any flaw in the reasoning above.

Very nice!!

Very nice!!
That works perfectly, because the x's cancel out when you add (100-x) and x
it certainly makes my solution look very cumbersome :-)

Hi Brent, the question below

Hi Brent, the question below has got me twisted. I'm kinda confused about using the same variable "x" for both boys and girls. Secondly, the proportions are not making sense to me either as to why you have "x" over for 100 for the girls in one class, but 150 - x over 150 for the other class. Maybe I'm not seeing this or something is so obvious that it is escaping me at the present.

Class B has 50% more students than class A. Number of girls in class A is equal to number of boys in class B. The percentage of girls is the same in both classes. What percentage of the student group are boys?
A. 25 %
B. 33 %
C. 40 %
D. 50 %
E. 60 %

Number of girls in class A is equal to number of boys in class B.
Let x = number of girls in class A
So, x = number of boys in class B

This ALSO means that 100-x = number of boys in class B
And 150-x = number of girls in class B

The percentage of girls is the same in both classes.
In other words, the proportion of girls is the same in both classes.
In class A, the proportion of girls to the TOTAL class A population = x/100
In class B, the proportion of girls to the TOTAL class B population = (150-x)/150
So, we can write: x/100 = (150-x)/150
Cross multiply: 150x = 100(150-x)
Simplify: 150x = 15000 - 200x
Add 200 x to both sides: 250x = 15000
Solve: x = 60

Everything looks perfect . .

Everything looks perfect . . . so far (but you're not done yet :-).

You correctly determined that x = 60

So, in class B (out of 150 students) there are 60 boys.
And, in class A (out of 100 students) there are 60 girls. From this, we can conclude that there are 40 boys in class A.

So, the TOTAL number of boys = 60 + 40 = 100

TOTAL number of students = 150 + 100 = 250

So, PERCENTAGE of boys = 100/250 = 40%

Here's my step-by-step solution: https://gmatclub.com/forum/class-b-has-50-more-students-than-class-a-num...

My Bad! The solved part was

My Bad! The solved part was copied from your example! I was just showing you where I was confused.

Sorry about that - I didn't

At the beginning of my solution, I decided to assign "nice values" to the number of students in each class, so that class B has 50% more students than class A. I chose 100 for the total population of class A, and 150 for the total population of class B.

ASIDE: I could have chosen other values that satisfy the condition that "class B has 50% more students than class A," but the values I chose are easy to work with.

We're also told that "the number of girls in class A is EQUAL to number of boys in class B.
So, I let x = the number of girls in class A AND x = number of boys in class B.

From here, we can determine the number of girls in class B.
Number of girls in class B = total class B population - number of boys
= 150 - x

At this point, we can use the information that says "The percentage of girls is the same in both classes."

This is the same as saying that "The FRACTION of girls is the same in both classes."

The FRACTION of girls in class A = x/100 (there are x girls in class A, with a total class A population of 100)

The FRACTION of girls in class B = (150-x)/150 (there are 150-x girls in class B, with a total class B population of 150)

Since the two fractions are equal, we get the equation x/100 = (150-x)/150

Does that help?

Cheers,
Brent

HI Brent, can you help me

Hi Brent, can you help me with this question:

The average age of 40 women decreases by 1/8th of a year when one of them whose age is 50 years is replaced by a new woman. Find the age of the new woman.

a)36 years
b)42 years
c)45 years
d)43 years

Let's assign a "nice" value

Let's assign a "nice" value to the INITIAL average age of the 40 women.
Let's say that the INITIAL average age is 50
This means the INITIAL SUM of the 40 ages = (40)(50) = 2000
NOTE: Saying that the average age is 50 is fine since we are still meeting the conditions set out in the question

"The average age decreases by 1/8th of a year when one of them whose age is 50 years is replaced by a new woman."
When we remove one 50-year-old, the SUM of the 39 ages = 2000 - 50 = 1950

Let x = age of new (replacement) woman
Once we add her to the group, we have 40 women again
So, the NEW SUM of the 40 ages = 1950 + x

We're told that the NEW average is 1/8 of a year less than the INITIAL average
The INITIAL average was 50
So, the NEW average must be 49 7/8 years

Now comes our equation...
(NEW SUM of the 40 ages)/40 = 49 7/8
We get: (1950 + x)/40 = 49 7/8
Multiply both sides by 40 to get: 1950 + x = 1995
Solve: x = 45

Cheers,
Brent

Thanks Brent - really clear

Thanks Brent - really clear and great explanation!

This is such a cruel Q!
C is an easy trap since it way more easier to solve. Statement 1 alone is about finding possibilities which could be a long process isn't it?

Is there a trick to solve statement 1?
Thanks!

It's a devious question!

It's a devious question!
You're right; listing all of the possibilities can be quite time-consuming. However, we can apply some number sense to eliminate a bunch of the possible outcomes.

To see what I mean, check out my full solution: https://gmatclub.com/forum/joanna-bought-only-0-15-stamps-and-0-29-stamp...

Cheers,
Brent

For the video above, why are

For the video above, why are we multiplying Liam (2g+5) by 3 (Zoe). Why are we NOT multiplying Zoe (10G-13) by 3? It says Zoe is 3 times the age of Liam? Shouldn't we multiply Zoe by 3?

You're referring to the

You're referring to the question at 2:50 in the above video.

This is a great example of why it's useful to start with a word equation.

GIVEN: Zoe is 3 times as old as Liam
Rewrite as: Zoe's age IS 3 times Liam's age
Rewrite as: Zoe's age = 3 times Liam's age
Rewrite as: Zoe's age = 3(Liam's age)

Now replace the ages with their algebraic expressions.
We get: 10G - 13 = 3(2G + 5)
Done!!

ASIDE: Our goal with these kinds of questions is to create an EQUATION. That is, we want to manipulate one quantity so that we get an EQUATION.
If you're not sure if your equation is correct, you can always test values along the way.

We're told that Zoe is 3 times as old as Liam
So, one possible scenario is that Liam is 2 years old, and Zoe is 6 years old.
How do we make their ages EQUAL?
Do we multiply Liam's by 3, or do we multiply Zoe's by 3?

Well, if we take the older child (Zoe, age 6) and multiply her age by 3, we get 18, which is not EQUAL to Liam's age.

Conversely, if we take the younger child (Liam, age 2) and multiply his age by 3, we get 6, which is EQUAL to Zoe's age.

Does that help?

Cheers,
Brent

Hi Brent,

Hi Brent,

I am struggling with this question: https://gmatclub.com/forum/a-grocery-store-sells-apples-by-the-pound-if-the-price-per-pound-is-i-258600.html
I don't understand why C-1= the increased price per pound. Why isn't it C+1?

Thank you!
BR Pia

Good catch! I've edited my response accordingly.

Cheers and thanks,
Brent

may i have your solution of

may i have your solution of this
https://gmatclub.com/forum/at-a-certain-picnic-each-of-the-guests-was-served-either-a-90254.html