Quantitative reasoning for my Law Aptitude test..

roel · 2007-11-26 12:23:51

Hi.. I am preparing for my Law aptitude test. Could you please help me answer the questions in my reviewer? I already have my own answers but i do not know if i answered it right. It is just because I am not good in math.. This is more than 90 questions i think. If it's ok, can I post 5 problems a day? please... This is the first five. Just tell me if i can post more next time.. Thanks...

DAY 1

1. If r≠0 and r=r-², what is the value of r?

2. Find the sum: 23 + 24 + 25 + ..........+ 70 + 71 + 72

(thanks to mathsyperson for helping me make the fractions...)

4. In a certain class, 1/2 of the male students and 2/3 of the female students read harry potter books. If there are 3/4 as many girls as boys in the class, what fraction of the entire class reads Harry Potter books?

5. The price of an item is increased by 25%. When the item was not sold, it is decreased by 50% of the increased price. If the item now costs P250, what is its original price?

If it's possible, please answer it in the simplest way since it's hard for me to understand complex solutions.. Thanks again.. I hope you'll allow me to post more next time.. Thanks..

Last edited by roel (2007-11-26 21:55:12)

mathsyperson · 2007-11-26 13:31:30

1. If r = r[sup]-2[/sup], then multiplying by r² gives that r³=1.
Hencely, r = 1. (And technically (-1±√3i)/2 as well, but I'm guessing you don't care about imaginary stuff)

2. This is equal to the 72nd triangular number, minus the 22nd one.
The nth triangular number is given by n(n+1)/2, and so your sum is (72*73)/2 - (22*23)/2 = 2375.

3. You can make fractions with the following code: (math)\frac{a}{b}(/math), but with [] instead of () around the math tags.

That would produce

, so then you just replace the a and b in each of the curly braces with whichever expression you want.

There's nothing wrong with just writing fractions in plaintext either, but if you do then make sure it's clear what you mean.
For example, does 2/3x mean (2/3)x or 2/(3x)?

4. If there are 3/4 as many girls as boys in the class, then 3/7 of the class are girls and 4/7 are boys.

Then, 2/3*3/7 = 2/7 of the class are girls who read Harry Potter, and 4/7*1/2 = 2/7 of the class are boys who read Harry Potter. Adding those up, 4/7 of all the class read Harry Potter.

5. Let's say that the initial price was x.
Then it was increased by 25%, and so the price became 5x/4.
It then dropped by 50%, to 5x/8.

We're told that this price is 250, and so we know that 5x/8 = 250.
Solve this: 5x/8 = 250 --> 5x = 2000 --> x = 400.

Therefore the original price was P400.

roel · 2007-11-26 15:01:37

thanks... I will print all the posts here in this thread.. I already have a reviewer.. ... can i post again tomorrow?

roel · 2007-11-26 15:20:36

mathsyperson wrote:

5. Let's say that the initial price was x.
Then it was increased by 25%, and so the price became 5x/4.
It then dropped by 50%, to 5x/8.
We're told that this price is 250, and so we know that 5x/8 = 250.
Solve this: 5x/8 = 250 --> 5x = 2000 --> x = 400.
Therefore the original price was P400.

where did you get the 5 in 5x/4? sorry.. again, i am not good in math..:(

bossk171 · 2007-11-26 16:49:15

25% = 1/4

100% = 1

100% + 25% = 1 + 1/4 = 4/4 + 1/4 = 5/4

You start with one x, then you add 25% of x to that, and that makes 125% of x.

roel · 2007-11-27 17:26:00

Thanks to those who answered..

here's for:

DAY 2

1. The denominator of a certain fraction is 3 less than twice the numerator. If 5 is added to both the numerator and the denominator, the denominator becomes 1 more than the numerator. Find the original fraction..

for numbers 2-3

A launches her first painting exhibit. One of her paintings was sold. That painting has a length of 15 inches shorter than thrice its width and a perimeter of 98 inches.

2. Find the length, in inches, of the painting.

3. If the man who bought the painting paid P 673,200.00 for the painting, how much did he pay for every square inch of the painting?

4. A's bookshelf includes books of P books, G books, and M books. If the ratio of P books to G books is 5:3 and P books to M books is 6:4, what is the ratio of the number of M books to the number of G books?

5. The average weight of ten people in an elevator is 145 pounds. If one person gets off the elevator, the average weight of the remaining people is 150 pounds. What is the weight, in pounds, of the person who gets off the elevator?

Thanks again for your help...

TheDude · 2007-11-28 01:14:13

1. We'll call the denominator d and the numberator n. We know that the denominator is 3 less than twice the numerator, which mathematically means d = 2n-3. We are also told that if 5 is added to both the numerator and denominator then the denominator becomes 1 more than the numerator, but this is the same as just saying that the denominator is 1 more than the numerator, which means d = 1+n. Since we know the value of d from the second equation we can substitute its value into the first equation to get d = 2n-3 ==> 1+n = 2n-3. You can now solve for both n and d.

2. We'll call the length l and the width w. We're told that the length is 15 inches less than 3 times the width, whicih means l = 3w-15. We also know that the perimeter of the painting is 98 inches. Assuming that the painting is rectangular this means that twice the width plus twice the length equals the perimeter, or 2l+2w = 98. Like before, we use substitution for l: 2(3w-15)+2w = 98 ==> 6w-30+2w = 98 ==> 8w = 128. From here you can solve for both w and l.

3. Since we now know the length and width of the painting, multiply them to find the area in square inches. Call the area A. Your answer is simply 673,200 divided by A.

4. Multiply the ratios to find a common value. Multiply the ratio of P to G by 6 to get 30:18, and multiply the ratio of P to M by 5 to get 30:20. This means that for every 30 P books you have 18 G books and 20 M books. Thus, the ratio of M to G is 20:18, which is equivalent to 10:9.

5. The average weight of 10 people is the sum of all their weights divided by 10. Since their average weight is 145 pounds, their total combined weight is 1450 pounds. When one person leaves that means that there are now 9 people with an average weight of 150 pounds, which means their total weight is 1350. Subtract the weights and you'll find that the person that left weighed 100 pounds.

mathsyperson · 2007-11-28 01:16:46

To answer 3. from day 1, you make the fractions' denominators all match and then get rid of them. You can do that because the denominators are all constant (there aren't any x's in them).

Also, for future reference, you don't have to put each fraction in separate [math] tags. You can make the code inside the tags as big as you want. The only problem with that is that if you make an error then it becomes harder to find it.

roel · 2007-11-28 03:41:52

TheDude wrote:

1. We'll call the denominator d and the numberator n. We know that the denominator is 3 less than twice the numerator, which mathematically means d = 2n-3. We are also told that if 5 is added to both the numerator and denominator then the denominator becomes 1 more than the numerator, but this is the same as just saying that the denominator is 1 more than the numerator, which means d = 1+n. Since we know the value of d from the second equation we can substitute its value into the first equation to get d = 2n-3 ==> 1+n = 2n-3. You can now solve for both n and d.

So, in this case, i can disregard the statement If 5 is added to both n and d.. right?
So, the answer here is

?

5. The average weight of 10 people is the sum of all their weights divided by 10. Since their average weight is 145 pounds, their total combined weight is 1450 pounds. When one person leaves that means that there are now 9 people with an average weight of 150 pounds, which means their total weight is 1350. Subtract the weights and you'll find that the person that left weighed 100 pounds.

If n here refers to the person who got off the elevator. then

1450-n=1350
n=100
right?

thanks a lot..

The rest, I tried to solve it here, with your answers as my guide. And I got it! thanks for your help...

Last edited by roel (2007-11-28 03:57:11)

roel · 2007-11-28 04:03:40

mathsyperson wrote:

To answer 3. from day 1, you make the fractions' denominators all match and then get rid of them. You can do that because the denominators are all constant (there aren't any x's in them).
Also, for future reference, you don't have to put each fraction in separate tags. You can make the code inside the tags as big as you want. The only problem with that is that if you make an error then it becomes harder to find it.

thanks.. but I am just a bit confused with solving fractions here.
Why is that in case we add for example

we retain the denominator? I.e.,

right?

But in this problem, why is the denominator 12 not retained? Sorry, i no longer remember what my teacher taught me...

thanks for your help..

NullRoot · 2007-11-28 04:21:07

Roel,

Absolutely right on the elevator one.

For this one:
1. The denominator of a certain fraction is 3 less than twice the numerator. If 5 is added to both the numerator and the denominator, the denominator becomes 1 more than the numerator. Find the original fraction..

Try writing out the statements in terms of equalities.

So:
The denominator of a certain fraction is 3 less than twice the numerator.
D = (2N)-3

If 5 is added to both the numerator and the denominator, the denominator becomes 1 more than the numerator.
D+5 = 1+(N+5)

The second one can be cleaned up. You don't need to add A+B+C in any specific order, so the parenthesis are unnecessary; I used them to better represent the sentence. So:
D+5 = 1+N+5

If you Subtract 5 from both sides, then it simplifies to: D = 1+N, as TheDude said. Because we know this, we can take the first equality and have only 1 variable in it. We do this by substitution:
1+N = 2N-3
Adding 3 to both sides:
4+N = 2N
Subtract 1N from both sides:
4 = N

Putting N back into D=(2N)-3, you work out that D does in fact equal 5, so your fraction is 4/5. Good job

As far as Mathsyperson's solution, he just did two steps at once. When he added it all up, it would have come out to:

If you multiply both sides by 12, then you will cancel out the divison and arrive at:

Good luck with your Law test. I have a friend who's studying Criminal Justice at the moment .

Last edited by NullRoot (2007-11-28 04:25:38)

mathsyperson · 2007-11-28 04:26:10

You have to retain the denominator in that case because you're finding the value of an expression.

When you have an equation, you can do whatever you want to it, as long as you do the same thing on both sides of the equals sign. So at that step I just multiplied everything by 12.

You could keep the denominator in if you wanted to, and that would give you

.

At that point you would have to multiply by 12 in order to get x on its own.

Edit: Gah, beaten to it. Ah well, we agree at least.

roel · 2007-11-28 14:00:25

NullRoot wrote:

Roel,
Absolutely right on the elevator one.
For this one:
1. The denominator of a certain fraction is 3 less than twice the numerator. If 5 is added to both the numerator and the denominator, the denominator becomes 1 more than the numerator. Find the original fraction..
Try writing out the statements in terms of equalities.
So:
The denominator of a certain fraction is 3 less than twice the numerator.
D = (2N)-3
If 5 is added to both the numerator and the denominator, the denominator becomes 1 more than the numerator.
D+5 = 1+(N+5)
The second one can be cleaned up. You don't need to add A+B+C in any specific order, so the parenthesis are unnecessary; I used them to better represent the sentence. So:
D+5 = 1+N+5
If you Subtract 5 from both sides, then it simplifies to: D = 1+N, as TheDude said. Because we know this, we can take the first equality and have only 1 variable in it. We do this by substitution:
1+N = 2N-3
Adding 3 to both sides:
4+N = 2N
Subtract 1N from both sides:
4 = N
Putting N back into D=(2N)-3, you work out that D does in fact equal 5, so your fraction is 4/5. Good job
As far as Mathsyperson's solution, he just did two steps at once. When he added it all up, it would have come out to:
If you multiply both sides by 12, then you will cancel out the divison and arrive at:
Good luck with your Law test. I have a friend who's studying Criminal Justice at the moment .

oh really?

Thanks a lot for helping me NullRoot. .

roel · 2007-11-28 14:04:12

mathsyperson wrote:

You have to retain the denominator in that case because you're finding the value of an expression.
When you have an equation, you can do whatever you want to it, as long as you do the same thing on both sides of the equals sign. So at that step I just multiplied everything by 12.
You could keep the denominator in if you wanted to, and that would give you
.
At that point you would have to multiply by 12 in order to get x on its own.
Edit: Gah, beaten to it. Ah well, we agree at least.

I see.. I got it.. It is an equation if we're looking for a certain value of say, X right? and it is an expression if we are solving for the entire fraction, say adding 3/4 + 1/2?

thanks a lot mathsyperson..

roel · 2007-11-28 15:12:17

Here's for:

DAY 3

1. An empty pool being filled with water at a constant rate takes 8 hours to fill

of its capacity. How much more time will it take to finish filling the pool?

2. A union contract specifies a 6% salary increase plus a P450 bonus for each employee. For a certain employee, this is equivalent to an 8% salary increase. what was the employee's salary before the contract?

3. Simplify

4. If

, find the value of x.

That's all for now so we can finish the first material.. Anyway, how can I make exponents here? i tried to copy the exponents above but everytime i use the [math] code, it won't appear..

Thanks again. i hope you won't get tired helping me...

Identity · 2007-11-28 17:35:32

DAY 3

1

The rate at which the pool is being filled is

. This represent the fraction filled every hour.
Since rate is constant, this is equal to , where x is the amount of time taken to fill the pool to the remaining 2/5th capacity.

So the remaining 2/5th of the pool is filled in 5 and 1/3 hours.

2
Let x be the employee's original salary.
The equation is:

So his salary is $22500

3

4 I use reciprocation twice here. It can be very useful.

Reciprocating (flipping both sides):

Subtracting 1 from both sides (leaving fractions on both sides so I can flip again and simplify things)

Reciprocation (flipping again)

Last edited by Identity (2007-11-28 20:51:33)

roel · 2007-11-28 19:42:51

Identity wrote:

DAY 3
4 I use reciprocation twice here. It can be very useful.

Thanks.. everything is clear to me.. Except no. 4..

How did you get this?

is there another way of solving this?

Thanks Identity..

roel · 2007-11-28 19:44:22

where can i find some problems that are almost similar with all these?

Identity · 2007-11-28 20:49:35

I edited my previous post explaining the steps

Also:

or if you prefer,

but this is basically where my first solution reached on its second line.

These types of problems pop up all over competition papers. There is an excellent website where you can download them.

http://www.math.ksu.edu/main/handbook/ProblemSets

You can find these types of problems normally in the first 10-20 questions of the AMC papers. Good thing is you get the fully-worked solutions too! Have fun

Last edited by Identity (2007-11-28 21:08:54)

roel · 2007-11-28 21:10:49

thanks identity.. i will post again some problems tomorrow. please don't get tired helping me. Meanwhile, i will have a practice with the problems in the site you posted.. thanks again..

roel · 2007-11-28 21:14:55

Identity wrote:

I edited my previous post explaining the steps
Also:
or if you prefer,
but this is basically where my first solution reached on its second line.
These types of problems pop up all over competition papers. There is an excellent website where you can download them.
http://www.math.ksu.edu/main/handbook/ProblemSets
You can find these types of problems normally in the first 10-20 questions of the AMC papers. Good thing is you get the fully-worked solutions too! Have fun

I think, ill go with the first one... thanks again...

roel · 2007-11-28 21:26:17

http://www.math.ksu.edu/main/handbook/ProblemSets

wew... i hope i wont be seeing questions in the LAT similar with those..

Identity · 2007-11-28 22:44:37

Hmmm, maybe they are too hard. But hey! If you can do some hard ones, then the easy ones will be easier. Anyway... I think some of AMC8 might be suitable, and I don't know many other places to find problem solving questions like these.

NullRoot · 2007-11-28 22:53:05

Identity has a very good point. Try out the hard ones as best you can and then check the solutions when you feel you're stuck or if you have an answer. It might put you in the right state of mind to deal with problems that are easier. There's some very good examples on there that get you thinking about how to turn a word problem into equations

roel · 2007-11-28 23:08:33

ahaha.. that's what I did a while ago... I only answered 5 correct answers I think.. ahaha. I need some break.. I will try n to solve those problems again tomorrow. Thanks again mates..

Math Is Fun Forum

#1 2007-11-26 12:23:51

Quantitative reasoning for my Law Aptitude test..

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Re: Quantitative reasoning for my Law Aptitude test..

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