Question: Nessa and Capri

Comment on Nessa and Capri

please solve with 2 variables if it is possable
greenlight-admin's picture

Sure thing.
Let N = # of CDs that Nessa has
Let C = # of CDs that Capri has

"The number of CDs that Nessa has is 14 less than twice the number of CDs that Capri has"
We can write: N = 2C - 14

"If Nessa gives 20% of her CDs to Capri, then Nessa and Capri will have the same number of CDs"
20% of Nessa's CDs = 0.2N
So, the # of CDs Nessa has = N - 0.2N
So, the # of CDs Capri has = C + 0.2N
So, we can write: N - 0.2N = C + 0.2N
Simplify: 0.8N = C + 0.2N
Subtract 0.2N from both sides to get: 0.6N = C

We now have two equations:
N = 2C - 14
0.6N = C

Take top equation and replace C with 0.6N to get: N = 2(0.6N) - 14
Simplify: N = 1.2N - 14
Subtract 1.2N from both sides: -0.2N = -14
Solve: N = 70

To solve for C, we'll use the fact that 0.6N = C
So, 0.6(70) = C
So, C = 42

TOTAL number of CDs they have = N + C = 70 + 42 = 112

Hi Brent

"The number of CDs that Nessa has is 14 less than twice the number of CDs that Capri has" We can write: N = 2C - 14

Will the Solution be same if I write: N= 14-2C

If NOT, why? And How can I know that it should be only 2C-14 and not 14-2C
greenlight-admin's picture

Let's take it step-by-step

If the number of CDs that Nessa has is TWICE the number of CDs that Capri has, we can write: N = 2C
If the number of CDs that Nessa has is 14 LESS THAN twice the number of CDs that Capri has, we write: N = 2C - 14

So, for example, if Capri has 10 CDs, then N = 2(10) - 14 = 6, which means Nessa has 6 CDs.
Likewise, if Capri has 25 CDs, then N = 2(25) - 14 = 36, which means Nessa has 36 CDs.
And if Capri has 43 CDs, then N = 2(43) - 14 = 72, which means Nessa has 72 CDs.

On the other hand, N = 14 - 2C represents a different situation altogether.
This equation suggests that the number of CDs Nessa has is equal to 14 minus twice the number of CDs Capri has.
So, for example, if Capri has 5 CDs, then N = 14 - 2(5) = 4, which means Nessa has 4 CDs.

In general, x - y is not the same as y - x (unless x = y, in which case both expressions evaluate to be 0)

Does that help?

Yes, May Be I have practice on framing the equation involving the subtractions. For All X-Y I'm framing opposite equations like Y-X. Though I'm doing that, I'm analyzing the end result and making them positive and proceeding further. For an example - In a age problem I get some Tommy's age =-8, that doesn't make any sense. So immediately, I'm force converting it to positive and proceeding with further calc. Anyhow, Thanks for helping me finding the mistake.
greenlight-admin's picture

There are definitely times in which accidentally writing x-y (instead of y-x) will yield the negative version of the correct answer. However, there will also be times when such a mistake will yield the wrong answer all together.

Another thing you can do is test the answers. You have one equation being N = 2C - 14. Each answer is the sum of the CDs, so N + C = (answer). Solve those systems of equations for N and C. Then see what 20% of N is. Add that to C and subtract it from N. If both answers are equal, you have the right answer. It's D.

C + N = 112
N = 2C - 14 (rewrite this as 2C - N = 14)

Now solve these:
C + N = 112
2C - N = 14
The N's cancel out. 3C = 126. C = 42.
Since 42 + N = 112, N=70

So we have N = 70 and C = 42.
20% of 70 is 14. 70 - 14 = 56. 42 + 14 = 56.
BOOM. There it is.

Thanks for the solution in the comments too! That's the way I had worked it out but got stuck on the last step. That was really helpful.

I have N=(2c-14)(.80) and N=1.20c. I set the equations equal to each other and got C=28. When I plugged that into N I got 42.
greenlight-admin's picture

There's a problem with your equation N=1.20c.
Just because Nessa gives 20% of her CDs to Capri, we can't conclude that Capri's CDs have increased by 20%.

For example, consider what happens if Nessa has 100 CDs and Capri has 5 CDs. This would mean that Nessa will give 20 of her CDs to Capri. So, Capri's CD collection increase from 5 CDs to 25 CDs, which represents a 400% increase (not a 20% increase).

This one is still easy enough to solve by testing answers.
Since N = 2c-14
and we're looking for
N + c
= 2c-14 + c
= 3c-14

Therefore the answer after adding 14 should be divisible by 3 and only one answer choice meets this condition.
greenlight-admin's picture

Very nice approach!!!

so when you multiply a side of an equation by the denominator of a fraction to get rid of the fraction, any part of the equation in parenthesis doesn't need to be multiplied?
greenlight-admin's picture

I'm not entirely sure which part of the solution you're referring to, but here are a few examples of eliminating fractions when dealing with equations:

Take the following equation: x/5 = 7
Multiply both sides by 5/1 to get: (5/1)(x/5) = (5/1)(7/1)
Expand: 5x/5 = 35/1
Rewrite the left side as follows: (5/5)(x/1) = 35/1
Since 5/5 = 1, we can write: (1)(x/1) = 35/1
And this is the same as: x = 35

Yes, this is a VERY convoluted way to solve a super easy equation. The important thing I want you to see is that the technique works because we end up with the fraction (5/5), which simplifies to be 1.

Here's another example: (2/3)(x + 1) = (1/6)(5x - 7)
Multiply both sides of the equation by 6 (aka 6/1) to get: (6/1)(2/3)(x + 1) = (6/1)(1/6)(5x - 7)
Multiply the first two fractions on each side of the equation: (12/3)(x + 1) = (6/6)(5x - 7)
Simplify the first fraction on either side: (4)(x + 1) = (1)(5x - 7)
Expand: 4x + 4 = 5x - 7
Solve: x = 11

For more information about simplifying equations involving fractions, you can watch: https://www.greenlighttestprep.com/module/gre-algebra-and-equation-solvi...

thanks

FYI, I realized after reading your comment and attempting this problem again what exactly it was that I didn't understand. I think my issue with eliminating fractions is not knowing what is in the numerator. In other words, what part of the expression is overtop of the denominator. In this case--1/5 (2c-14)--I didn't realize that (2c-14) was also over 5, I thought just the 1 was. So when I multiplied by 5, I was eliminating the fraction but ending up with 10c-70.

Anyway, just wanted to share with you because I thought it might be useful for you to know more specifics about what I was missing. I am understanding it better now and your comment helped a lot!
greenlight-admin's picture

That's great that you were able to identify the source of your difficulties. Your new understanding will help a lot with other similar questions.

I don't understand if they end up with the same number of CDs and we set the expressions equal to each other, then why isn't the answer 84 or 2x42? Why did you plug C back into the original equations to arrive at 112?
greenlight-admin's picture

Great question!

This all comes down to how we defined our variables.
In this case, C = the number of CDs that Capri PRESENTLY owns.

Later, after some CDs are exchanged (resulting in Capri and Nessa having the same number of CDs), we determine that C = 42.

That is, Capri PRESENTLY owns 42 CDs, whereas the information Capri and Nessa having the same number of CDs is a HYPOTHETICAL EVENT that has not actually happened.

So, once we know that Capri PRESENTLY owns 42 CDs, we can use the other piece of information regarding the PRESENT situation (i.e., "The number of CDs Nessa (PRESENTLY) has is 14 less than twice the number of CDs Capri has") to determine the number of CDs Nessa PRESENTLY has.

Does that help?

Hi Brent, I have an algebra question: when do we multiply by the opoosite to get 1 and when do we multiple by the denominator to make it an integer? Here we can't do the former because then the equation wouldn't work right - but when do we generally do want to multiply by the opposite (so like 3/5 would be multiplied by 5/3 e.g.). Thanks hope it makes sense my question.
greenlight-admin's picture

If you have TWO OR MORE fractions with different numerators, we can eliminate all fractions by multiplying by the least common multiple of all the denominators.
For example: x/3 + 5/6 = x/4 + 17/6
Multiply both sides by 12 to get: 4x + 10 = 3x + 34
Solve: x = 24

If you have a fraction with just ONE fraction in front of a variable, then you can multiply both sides of the equation by the reciprocal.
For example: (2/3)x = 18
Multiply by 3/2 to get: x = (18)(3/2) = 27

Does that help?

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