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1:if x and 10 are relatively prime natural numbers, then x [#permalink]
17 Sep 2008, 03:00

1:if x and 10 are relatively prime natural numbers, then x could be a multiple of a: 9 b: 18 c:4 d:25 e:14

2:the area of ADE is 12 square units. If B is the midpoint of AD and C is the midpoint of AE, what is the area of ABC. a:2 s/u b:3 s/u c:3.5 s/u d:4 square units e: 6 square units

I know the answer to these questions; however, i do not understand how to solve the problem. If someone could explain them in details, i would really appreciate it. Thank you very much.

Re: 2 math questions: [#permalink]
17 Sep 2008, 12:15

1

This post received KUDOS

In response to question 2 (is the answer 3?). If that is the right answer, I drew a triangle A, top D, and E with B as AD bisector and C as base bisector then formed the rest of the internal triangle that began with the line BC. So now we have an upside down triangle inscribed in the right side up triangle, forming 4 separate small triangles. In my eyes...ABC looks like 1/4 of the area of 12..or 3. If that's not the answer...I'm stumped too.

Re: 2 math questions: [#permalink]
17 Sep 2008, 12:49

I think relatively prime natural numbers are numbers that do not have any common prime factor.. So 1st question, answer should be A ( 9 = 3*3 , 3 is not a factor of 10=5*2 )

For the second one ABC and ADE are similar triangles.. so there sides are in the same ration.. i.e. AB/AD=AC/AE=BC/DE = > the height of the triangle ABC will be half the height of ADE ..

so area of ABC= 1/2* BC* height of ABC = 1/2* DE/2* (height of ADE)/2 = 1/4*12= 3 square units

If what you say is true...and X and 10 have no similar prime factors, then how can the answer be 9? The question asks which # can X be a multiple of. If X is a multiple of 9, then it can be 9 and then the disimilar factorization holds true but what if X is 18 (a multiple of 9)..then 18 and 10 have 2 as a common prime factor. I don't understand. Unless relatively prime natural numbers only include common even prime factorization? Help! Now it's become my question..thanks

Re: 2 math questions: [#permalink]
29 Sep 2008, 12:31

[color=#BF0000]you may have received the answer to these problems by now. I still want to share with you what I think, let me know what the solution is.

For problem 1

Prime numbers are numbers that can divide into 2 with a remender of 1. I think that the answer to that problem is A

Re: 2 math questions: [#permalink]
29 Sep 2008, 22:36

Nice explanation +1

icandy wrote:

charles77479a wrote:

1:if x and 10 are relatively prime natural numbers, then x could be a multiple of a: 9 b: 18 c:4 d:25 e:14

Relative prime numbers are numbers which have 1 as their Greatest common factor.

look at the Q, it says could be. lets look for which cannot be

18, 10 have 2 as a common divisor which is > 1

(4,10) (25,10) (14,10) all have divisors such as 2, 5 which are greater than 1. If the numbers themselves have GCF's other than 1, their multiples also will have GCF other than 1.

(9,10) have their greatest common factor as 1. even though (18,10) have 2 as common factor, 9 still serves our answer "could be"

2:the area of ADE is 12 square units. If B is the midpoint of AD and C is the midpoint of AE, what is the area of ABC. a:2 s/u b:3 s/u c:3.5 s/u d:4 square units e: 6 square units

Once we figure out that these triangles are similar triangles, we need to

Area of large triangle = (s ^ 2) * area of small triangle

where s is similarity ratio. We see that each side of big triangle is twice the small triangle

12 = 4 * area of small triangle -> 3

I know the answer to these questions; however, i do not understand how to solve the problem. If someone could explain them in details, i would really appreciate it. Thank you very much.

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