Problem Solving

If p is the perimeter of rectange Q, what is the value of p?

1) Each diagonal of rectangle Q has length 10.

2) The area of Q is 48.

I figured out that both (1) and (2) were insufficient on their own. After I combined the two peices of evidence I was able to get the formula down to the quadratic:

0 = (y^4) - 100(y^2) + 2304

Further, the factorization would look something like:

0 = (y^2 - ??) (y^2 - ??)

My question is what is the fastest way when you have a number like 2304 to pull the numbers to the factorization. Trial and error was my approach, but it simply takes too long.

Any help is greatly appreciated.

Thanks,

Jared

Answer is C, by deduction. The rectangle forms 2 right triangles, with a hypotenuse of 10.
The are of Q is 48. That means that L*W = 48. You have to recognize that this is a 3-4-5 triangle, so the only possibility is 6 * 8 = 48.

So 6^2 + 8^2 = 10^2.

Basically, whenever you see a triangle with a hypotenuse of 5 or 10...think 3-4-5 triangle.

Remember, that the GMAT will never ask you to factor something ridiculous like 2304. Very few people can do something like that. So try to look for another way.

That makes sense. Thanks a lot!

Haha, I don't know why I didn't do this earlier but there is another way I could of figured it out.

If I treat the formula as such:

(y^2 - a)(y^2 - b)

And I know that

a + b = 100

and

ab = 2304

I could solve solve the two equations and figure it out, but your way was definitely the quickest.

student22 wrote:Answer is C, by deduction. The rectangle forms 2 right triangles, with a hypotenuse of 10.
The are of Q is 48. That means that L*W = 48. You have to recognize that this is a 3-4-5 triangle, so the only possibility is 6 * 8 = 48.

So 6^2 + 8^2 = 10^2.

Basically, whenever you see a triangle with a hypotenuse of 5 or 10...think 3-4-5 triangle.

Remember, that the GMAT will never ask you to factor something ridiculous like 2304. Very few people can do something like that. So try to look for another way.

Is it not A???

gmatmachoman wrote:

student22 wrote:Answer is C, by deduction. The rectangle forms 2 right triangles, with a hypotenuse of 10.
The are of Q is 48. That means that L*W = 48. You have to recognize that this is a 3-4-5 triangle, so the only possibility is 6 * 8 = 48.

So 6^2 + 8^2 = 10^2.

Basically, whenever you see a triangle with a hypotenuse of 5 or 10...think 3-4-5 triangle.

Remember, that the GMAT will never ask you to factor something ridiculous like 2304. Very few people can do something like that. So try to look for another way.

Is it not A???

we cannot conclude from A that the triangle is a 3-4-5 triangle because the sides can be any thing as long as sum of their squares add up to 100 ex square of the sides can be 50,50 or 60,40 or 70,30 etc .this is why we need the next statement.

liferocks wrote:

gmatmachoman wrote:

student22 wrote:Answer is C, by deduction. The rectangle forms 2 right triangles, with a hypotenuse of 10.
The are of Q is 48. That means that L*W = 48. You have to recognize that this is a 3-4-5 triangle, so the only possibility is 6 * 8 = 48.

So 6^2 + 8^2 = 10^2.

Basically, whenever you see a triangle with a hypotenuse of 5 or 10...think 3-4-5 triangle.

Remember, that the GMAT will never ask you to factor something ridiculous like 2304. Very few people can do something like that. So try to look for another way.

Is it not A???

we cannot conclude from A that the triangle is a 3-4-5 triangle because the sides can be any thing as long as sum of their squares add up to 100 ex square of the sides can be 50,50 or 60,40 or 70,30 etc .this is why we need the next statement.

U r right in saying its not A.But explanation has some glitch!

Sides can be sqrt(50) & sqrt (50). and NOT (50,50).

If the perimeter would have been given as a integer, then A would be suffice..what say???

gmatmachoman wrote:

liferocks wrote:

gmatmachoman wrote:

student22 wrote:Answer is C, by deduction. The rectangle forms 2 right triangles, with a hypotenuse of 10.
The are of Q is 48. That means that L*W = 48. You have to recognize that this is a 3-4-5 triangle, so the only possibility is 6 * 8 = 48.

So 6^2 + 8^2 = 10^2.

Basically, whenever you see a triangle with a hypotenuse of 5 or 10...think 3-4-5 triangle.

Remember, that the GMAT will never ask you to factor something ridiculous like 2304. Very few people can do something like that. So try to look for another way.

Is it not A???

we cannot conclude from A that the triangle is a 3-4-5 triangle because the sides can be any thing as long as sum of their squares add up to 100 ex square of the sides can be 50,50 or 60,40 or 70,30 etc .this is why we need the next statement.

U r right in saying its not A.But explanation has some glitch!

Sides can be sqrt(50) & sqrt (50). and NOT (50,50).

If the perimeter would have been given as a integer, then A would be suffice..what say???

I have mentioned that square of the sides in the explanation..agreed that if it is mentioned that side s are integer then ans would have been A as no other Pythagorean triplet has 5 as hypotenuse

Ans is C.

if length is L and breadth is B, Area = L*B = 48 and
Diagonal^2 = L^2 + B^2 = 100

Since we have to find the perimeter = 2(L+B)

L^2 + B^2 + 2L*B = 100 + 2*48 = 196
(L+B)^2 = 196
(L+B) = 14 (Since its a rectangle, we will consider the positive root)
Perimeter = 2*(L+B) = 28

However, this a DS question, we need not resolve the complete equation. We have 2 variables and 2 equations, that should be sufficient to pick the answer as C.

sk818020 wrote:If p is the perimeter of rectange Q, what is the value of p?

1) Each diagonal of rectangle Q has length 10.

2) The area of Q is 48.

I figured out that both (1) and (2) were insufficient on their own. After I combined the two peices of evidence I was able to get the formula down to the quadratic:

0 = (y^4) - 100(y^2) + 2304

Further, the factorization would look something like:

0 = (y^2 - ??) (y^2 - ??)

My question is what is the fastest way when you have a number like 2304 to pull the numbers to the factorization. Trial and error was my approach, but it simply takes too long.

Any help is greatly appreciated.

Thanks,

Jared

An approach in which minimal calculations are required :

given a rectangle, say of length l and breadth b. with statements, can we find 2(l+b).

Statement 1 :

It tells us that l^2 + b^2 = 100.

Not sufficient.

Statement 2 :

it tells us that l*b = 48.

Not sufficient.


Combo:

we know that (l+b)^2 = l^2 + b^2 + 2 * l * b

we have all values in RHS and hence, we can find (l+b), which will give us the perimeter.

therefore, C will be the answer.

liferocks wrote:

gmatmachoman wrote:

student22 wrote:Answer is C, by deduction. The rectangle forms 2 right triangles, with a hypotenuse of 10.
The are of Q is 48. That means that L*W = 48. You have to recognize that this is a 3-4-5 triangle, so the only possibility is 6 * 8 = 48.

So 6^2 + 8^2 = 10^2.

Basically, whenever you see a triangle with a hypotenuse of 5 or 10...think 3-4-5 triangle.

Remember, that the GMAT will never ask you to factor something ridiculous like 2304. Very few people can do something like that. So try to look for another way.

Is it not A???

we cannot conclude from A that the triangle is a 3-4-5 triangle because the sides can be any thing as long as sum of their squares add up to 100 ex square of the sides can be 50,50 or 60,40 or 70,30 etc .this is why we need the next statement.

I didn't say it was A. I said it was C. I couldn't conclude anything from statement A.

But once I knew that the sides had to multiply to 48. The ONLY combination that's possible is 6 * 8 = 48. So, it's a 3-4-5 triangle 6-8-10.

I'm sorry I'm still not seeing how this is not answer "A". I understand the logic at arriving at answer "C", I just don't understand why you NEED to combine statements "1" and "2", contradicts my entire understanding of Data Sufficiency logic.

A rectangle is comprised of 4 right angles, no?

So ultimately the "diagonal" represents the hypotenuse forming two right triangles, no?

Can you form a right triangle with a hypotenuse of 10 with any other legs besides 6 and 8? Or do I have that wrong?

(pythagorean triplet (3, 4, 5) , (6, 8, 10))

repeat post

kelleygrad05 wrote:
Can you form a right triangle with a hypotenuse of 10 with any other legs besides 6 and 8? Or do I have that wrong?

(pythagorean triplet (3, 4, 5) , (6, 8, 10))

Kelley, your assumption is that you must take integer numbers as the values for side length or breadth.

The problem or the statement does not explicitly say that the sides have to be integers.

You may have infinite combinations of (l,b) such that l^2+b^2=100.

for distinct positive rational numbers represented by l, we may keep getting distinct values of b = sqrt(100-l^2) as long as b is also a positive rational

I am also confused with this question. In statement a) it is said, that both diagonals are equal. I looked into my mathbook and there is written: if it is a rectangle and both diagonals are equal, they bisect each other. With this knowledge we know two sides of the triangle and can solve with pythargoras theorem. Or is there any rectangle, where both diagonals are equal but DONT bisect ???