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## Thanks for the first round, help with Gmat prep q's!

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c210 Just gettin' started!
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Thanks for the first round, help with Gmat prep q's! Sun Apr 15, 2012 12:37 am
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Hey Guys and Gals,

everyone was really helpful when i first posted some questions regarding gmat prep q's and I really appreciate it. Here are some more, I hope you can take the time to help me figure it out!

1. If ax+b=0 is X>0?

1) A+b>0
2) A-b >0

I got this problem wrong, but the main issue is,what is the quickest easiest way to solve these type of problems? and whats the solution for this one

2. The number X and Y are not integers. the value of X is closest to which integer?

1) 4 is the integer closets to X + Y
2) 1 is the integer closest to X-Y

I have the same issue with this problem as I don't know the best way to approach it.

3. The sum of the first 50 positive integers is 2,550. What its he some of the even integers from 102-200 inclusive?

answer is 7550, and i have no idea how they came about it.

4. A certain circular area has its center at point P and radius 4, and the points X and Y lie int eh same plane as the circular area. Does point Y lie outside the circular area?

1) the distance between point P and point X is 4.5
2) the distance between point X and Y is 9

answer is C but why is it not A? couldn't you solve it already know that the X coordinate is outside of the radius?

5. if mv<pv<0 is V>0?
1)m<P
2)m<0

answer is D, why? how does M<p solve it? when you take into account M<p are you also taking into account that both are <0 ?

6. if x<0 then Sqrt(-x|x|) is? answer is -x...but shouldn't it be positive x?

7. if an equilateral triangle with side T and square with sides s had the same area, what is the ratio of T: S?

no idea why

8. At the bakery lew spent a total of 6\$ for one kind of cupcake and one kind of doughnut. How many donuts did he buy?
1) price of 2 doughty was \$.10 less than 3 cupcakes
2) average price of 1 doughnut and 1 cupcake was \$.035

answer is E, but couldn't you solve it with C ( since 2 equations and 2 unknowns are produced? )

9. A certain list consists of several different integers. is the product of all the integers in the list positive?

1) the product of the greatest and smaller of the integers in the list is positive
2) there is an even number of integers in the list

answer is C, wheres the logic in this?

and FINALLY

10) the function F is defined for all positive integers N by the following rule : F(N) is the number of positive integers each of which is less than N and has no positive factor in common with N other than 1. if P is any prime number then F(p) =?

a) p-1
b) p-2
c) (p+1)/2
d (p-1)/2
E) 2

i got the answer A correctly but was guessing, whats the smartest way to solve this problem and these kinds of problems?

thank you all so very much! i really appreciate the help!

best,

c

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Birottam Dutta GMAT Destroyer!
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Sun Apr 15, 2012 2:35 am
One post at a time would be appreciated!

Birottam Dutta GMAT Destroyer!
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Sun Apr 15, 2012 2:35 am
One post at a time would be appreciated!

c210 Just gettin' started!
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Sun Apr 15, 2012 9:35 pm
oh do you mean its easier to post 1 problem at a time per post? Sorry for the inconvenience! and thank you for the help in advance!

best,

c

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Anurag@Gurome GMAT Instructor
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Sun Apr 15, 2012 9:47 pm
Quote:
1. If ax+b=0 is X>0?

1) A+b>0
2) A-b >0

I got this problem wrong, but the main issue is,what is the quickest easiest way to solve these type of problems? and whats the solution for this one
Yes, it is definitely easier to reply for 1 query at a time.

Anyway, it is given that ax + b = 0 implies b = -ax
Question: Is x > 0?

(1) a + b > 0
a - ax > 0 (Since b = -ax)
a(1 - x) > 0 implies either a > 0 and (1 - x) > 0 implies x < 1 OR a < 0 and (1 - x) < 0 implies x > 1.

(2) a - b > 0
a - (-ax) > 0 (Since b = -ax)
a + ax > 0
a(1 + x) > 0 implies either a > 0 and (1 + x) > 0 implies x > -1 OR a < 0 and (1 + x) < 0 implies x < -1

Combining (1) and (2), a + b > 0 and a - b > 0
a + b + a - b > 0
2a > 0
a > 0 which implies x < 1. Also, x > -1
We get, -1 < x < 1, which implies x may or may not > 0; NOT sufficient.

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Thanked by: c210

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Anurag@Gurome GMAT Instructor
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Sun Apr 15, 2012 9:52 pm
Quote:
2. The number X and Y are not integers. the value of X is closest to which integer?

1) 4 is the integer closets to X + Y
2) 1 is the integer closest to X-Y

I have the same issue with this problem as I don't know the best way to approach it.
(1) 4 is the integer closets to X + Y implies 3.5 < (X + Y) < 4.5; NOT sufficient.
(2) 1 is the integer closest to X - Y implies 0.5 < (X - Y) < 1.5; NOT sufficient.

Combining (1) and (2), we add the two inequalities in statements 1 and 2,
(3.5 + 0.5) < (X + Y) + (X - Y) < (4.5 + 1.5)
4 < 2X < 6
2 < X < 3, which implies X can be closer to 2 as well as 3, as if X = 2.2, then it is closer to 2 and if X = 2.9 then X is closer to 3; NOT sufficient.

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c210 Just gettin' started!
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Sun Apr 15, 2012 9:52 pm
thank you! i totally approached that problem wrong and did a plug in of test numbers method

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Anurag@Gurome GMAT Instructor
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Sun Apr 15, 2012 9:53 pm
c210 wrote:
thank you! i totally approached that problem wrong and did a plug in of test numbers method
You are welcome.

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Anurag@Gurome GMAT Instructor
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Sun Apr 15, 2012 9:55 pm
Quote:
3. The sum of the first 50 positive integers is 2,550. What its he some of the even integers from 102-200 inclusive?

answer is 7550, and i have no idea how they came about it.
We have to find 102 + 104 + 106 +....+ 200 = (100 + 2) + (100 + 4) + (100 + 6) +....+ (100 + 100)
= (100 + 100 + 100 +...+ 100) + (2 + 4 + 6 +...+ 100)

It is given that 2 + 4 + 6 + .... + 100 = 2,550

Now 100 will occur 50 times, so the above sum can be written = (50 X 100) + 2,550 = 5,000 + 2,550 = 7,550

So, the correct answer is 7,550.

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Anurag Mairal, Ph.D., MBA
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Anurag@Gurome GMAT Instructor
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Sun Apr 15, 2012 10:01 pm
Quote:
4. A certain circular area has its center at point P and radius 4, and the points X and Y lie int eh same plane as the circular area. Does point Y lie outside the circular area?

1) the distance between point P and point X is 4.5
2) the distance between point X and Y is 9
(1) The distance between point P and point X is 4.5 gives no info on Y; NOT sufficient.

(2) The distance between point X and Y is 9 but there is no relationship between the ine segment and circle; NOT sufficient.

Combining (1) and (2), point X lies outside the circle (as the radius of circle is 4).
Now the closest point of the circle to the point X = 4.5 - 4 = 0.5
Farthest point = 0.5 + 8 = 8.5
Since XY = 9, so Y must be outside the circle: SUFFICIENT.

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Anurag@Gurome GMAT Instructor
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Sun Apr 15, 2012 10:07 pm
Quote:
5. if mv<pv<0 is V>0?
1)m<P
2)m<0

answer is D, why? how does M<p solve it? when you take into account M<p are you also taking into account that both are <0 ?
We are given that 1) mv < pv 2) mv < 0 3) pv <0
mv < pv implies (m - p)v < 0, which means either (m - p) > 0 and v < 0 or (m - p) < 0 and v > 0, which means either m > p and v < 0 or m < p and v > 0.

(1) m < p clearly answer the question (see the above steps). The answer to the question is v > 0; SUFFICIENT.

Also, mv < 0 implies that either m > 0 and v < 0 or m < 0 and v > 0. From here we can clearly say that statement 2 alone is SUFFICIENT since statement 2 says that m < 0. The answer is v > 0.

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Anurag@Gurome GMAT Instructor
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Sun Apr 15, 2012 10:10 pm
Quote:
6. if x<0 then Sqrt(-x|x|) is? answer is -x...but shouldn't it be positive x?
Note that √(x²) = |x|, which means that the square root cannot give a negative result.
So, √(x²) ≥ 0
Now if x = 5, then √(x²) = √25 = 5 = x = positive.
If x = -5, then √(x²) = √25 = 5 = -x = positive.

So, we can say that √(x²) = x if x ≥ 0.
and √(x²) = -x, if x < 0.

√(-x *|x|) = √(-x * -x) = √(x²) = |x| = -x

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Anurag@Gurome GMAT Instructor
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Sun Apr 15, 2012 10:23 pm
Quote:
7. if an equilateral triangle with side T and square with sides s had the same area, what is the ratio of T: S?

no idea why
Area of equilateral triangle, with each of the sides, t:

By Pythagoras Theorem, h² = t² - (t/2)² = t² - t²/4 = 3t²/4
h² = 3t²/4
h = t√3/2
Therefore, area of equilateral triangle = (1/2) * base * height = (1/2) * t * t√3/2 = t²√3/4

Area of square, with each side, s = s²

Now area of equilateral triangle and square are the same, so t²√3/4 = s²
So, t²/s² = 4/√3 or 4/3^(1/2)
t/s = 2/[3^(1/2)]^(1/2) = 2/3^(1/4)

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Anurag@Gurome GMAT Instructor
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Sun Apr 15, 2012 10:55 pm
Quote:
9. A certain list consists of several different integers. is the product of all the integers in the list positive?

1) the product of the greatest and smaller of the integers in the list is positive
2) there is an even number of integers in the list

answer is C, wheres the logic in this?
1st statement implies that either both of them are positive or both of them negative. It means all the integers between largest and positive integers should also have the same sign. So, either every integer in the list is positive, or every integer in the list is negative.
So, (1) is NOT SUFFICIENT.

2nd statement implies that product of integers is either positive or negative.
So, (2) is NOT SUFFICIENT.

Combining (1) & (2), the set of integers are either all positive and even or negative and even, which implies that the product will always be positive.

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Anurag Mairal, Ph.D., MBA
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Anurag@Gurome GMAT Instructor
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Sun Apr 15, 2012 10:56 pm
Quote:
10) the function F is defined for all positive integers N by the following rule : F(N) is the number of positive integers each of which is less than N and has no positive factor in common with N other than 1. if P is any prime number then F(p) =?

a) p-1
b) p-2
c) (p+1)/2
d (p-1)/2
E) 2

i got the answer A correctly but was guessing, whats the smartest way to solve this problem and these kinds of problems?

thank you all so very much! i really appreciate the help!
Tricky solution:

Let us take p = 2 (smallest prime)
Now number of positive integers less than p and has no common factor with p other than 1 is 1. So f(2) = 1

Only option A satisfies this result.

Mathematical Approach:

Note that a prime number will have common factors other than 1 only with its multiples like p², p³ etc. As p is always greater than 1, all multiples of p are greater than p. Hence, none of the integers less than p will have any common factor with p.

Thus, f(p) = Number of positive integers less than p = (p - 1)

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