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# First post, so I'm not sure if I needed to post each

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Joined: 03 Nov 2010
Posts: 13
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First post, so I'm not sure if I needed to post each [#permalink]

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03 Nov 2010, 14:56
00:00

Difficulty:

(N/A)

Question Stats:

33% (02:47) correct 67% (01:44) wrong based on 4 sessions

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First post, so I'm not sure if I needed to post each question as its own topic. These were in the GMAT software, but unfortunately didn't have the explanations for the answers.

---
DS

Are X > 0 & Y > 0
1) 2X - 2Y = 1
2) (X/Y) > 0

---

Is a(K) factor of b(M). K & M are exponents of a, b
1) A is a factor of b
2) K < M

---

For which of the following functions, F(A + B) = F(A) + F(B) for all positive numbers A & B
A) F(X) = x2
B) F(X) = x + 1
C) F(X) = SquareRoot X
D) F(X) = 2/X
E) F(X) = -3X

Thanks!
Kevin
Math Expert
Joined: 02 Sep 2009
Posts: 34438
Followers: 6257

Kudos [?]: 79467 [0], given: 10018

Re: DS/PS Questions from GMAT software [#permalink]

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03 Nov 2010, 15:16
KGG88 wrote:
First post, so I'm not sure if I needed to post each question as its own topic. These were in the GMAT software, but unfortunately didn't have the explanations for the answers.

---
DS

Are X > 0 & Y > 0
1) 2X - 2Y = 1
2) (X/Y) > 1

---

Is a(K) factor of b(M). K & M are exponents of a, b
1) A is a factor of b
2) K < M

---

For which of the following functions, F(A + B) = F(A) + F(B) for all positive numbers A & B
A) F(X) = x2
B) F(X) = x + 1
C) F(X) = SquareRoot X
D) F(X) = 2/X
E) F(X) = -3X

Thanks!
Kevin

Hi Kevin! Welcome to Gmat Club.

Yes, you should post one question per topic.

So make sure you type the question in exactly as it was stated from the source.

Your first 2 problems are DS questions and the third one is PS question:
Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/
Please post DS questions in the DS subforum: gmat-data-sufficiency-ds-141/

As for your questions:

1. Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1

(1) 2x-2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x-2y=1 --> y=x-1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient.

(2) x/y>1 --> x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.

(1)+(2) Again it can be done with different approaches. You should just find the one which is the less time-consuming and comfortable for you personally.

One of the approaches:
$$2x-2y=1$$ --> $$x=y+\frac{1}{2}$$
$$\frac{x}{y}>1$$ --> $$\frac{x-y}{y}>0$$ --> substitute x --> $$\frac{1}{y}>0$$ --> $$y$$ is positive, and as $$x=y+\frac{1}{2}$$, $$x$$ is positive too. Sufficient.

Discussed here: ds1-93964.html?hilit=number%20plugging%20consider%20approaches and here: inequality-and-absolute-value-questions-from-my-collection-86939.html

2. If a, b, k, and m are positive integers, is a^k factor of b^m?
(1) a is a factor of b.
(2) k = m[/quote]

Question: is $$\frac{b^m}{a^k}=integer$$?

(1) a is a factor of b --> $$ax=b$$, where $$x$$ is a positive integer --> is $$\frac{a^m*x^m}{a^k}=a^{m-k}*x^m=integer$$, now if $$m-k\geq{0}$$ then $$a^{m-k}*x^m$$ will be an integer BUT if $$m-k<{0}$$ then $$a^{m-k}*x^m$$ may NOT be an integer (well it will be an integer if $$a=1$$, but if $$a\neq{1}$$, then no). Not sufficient.

(2) k = m --> is $$\frac{b^k}{a^k}=(\frac{b}{a})^k=integer$$? --> if $$\frac{b}{a}=integer$$ then the answer would be YES, but if $$\frac{b}{a}\neq{integer}$$ then the answer would be NO. Not sufficient.

(1)+(2) $$ax=b$$ and $$k=m$$ --> $$\frac{b^m}{a^k}=\frac{a^k*x^k}{a^k}=x^k=integer$$. Sufficient.

Discussed here: factoring-concept-99665.html

3. For which of the following functions is f(a+b)= f(a)+f(b) for all positive numbers a and b?
A. $$f(x)=x^2$$
B. $$f(x)= x+1$$
C. $$f(x) = \sqrt{x}$$
D. $$f(x)=\frac{2}{x}$$
E. $$f(x) = -3x$$

A. $$f(a+b)=(a+b)^2=a^2+2ab+b^2\neq{f(a)+f(b)}=a^2+b^2$$

B. $$f(a+b)=(a+b)+1\neq{f(a)+f(b)}=a+1+b+1$$

C. $$f(a+b)=\sqrt{a+b}\neq{f(a)+f(b)}=\sqrt{a}+\sqrt{b}$$.

D. $$f(a+b)=\frac{2}{a+b}\neq{f(a)+f(b)}=\frac{2}{a}+\frac{2}{b}$$.

E. $$f(a+b)=-3(a+b)=-3a-3b=f(a)+f(b)=-3a-3b$$. Correct.

OR, as f(a+b)= f(a)+f(b) must be true for all positive numbers a and b, then you can randomly pick particular values of a and b and check for them:

For example: $$a=2$$ and $$b=3$$

A. $$f(a + b) = f(5) = 5^2 = 25\neq{f(a) + f(b) = f(2) + f(3) = 2^2 + 3^2 = 13}$$

B. $$f(a + b) = f(5) = 5 + 1 = 6\neq{f(a) + f(b) = f(2) + f(3) = (2 + 1) + (3 + 1) = 7}$$

C. $$f(a + b)=f(5)=\sqrt{5}\neq{f(a) + f(b)=f(2)+f(3)=\sqrt{2}+\sqrt{3}}$$

D. $$f(a + b)=f(5)=\frac{2}{5}\neq{f(a)+f(b)=f(2)+f(3)=\frac{2}{2}+\frac{2}{3}=\frac{5}{3}}$$

E. $$f(a + b)=f(5)=-3*(5) =-15=f(a)+f(b)=f(2)+f(3)=-3*(2)-3*(3)=-15$$. Correct.

It might happen that for some choices of a and b other options may be "correct" as well. If this happens just pick some other numbers and check again these "correct" options only.

Discussed here: functions-problem-need-help-93184.html?hilit=following%20functions%20positive%20numbers#p717196

Hope it helps.
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Re: DS/PS Questions from GMAT software [#permalink]

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03 Nov 2010, 19:05
Bunuel has already mentioned some solid approaches above. Let me add a couple more.

Are X > 0 & Y > 0
1) 2X - 2Y = 1
2) (X/Y) > 1

(BTW, if the second statement were (X/Y) > 0, then the answer would be E)

Ques: Are X and Y positive?
Statement (I): X - Y = 1/2. But this could be true for X and Y both positive e.g. X = 2.5, Y = 2, for X positive (=0.5), and Y equal to 0, for X = 0 and Y = -0.5, for X and Y both negative e.g. X = -2, Y = -2.5. Hence not sufficient.
Statement (II): X/Y > 1 means X and Y are either both positive or both negative. It also means that absolute value of X is greater than absolute value of Y. e.g. X = 2.5, Y = 2; or X = -2.5, Y = -2. This alone is also not sufficient.
Taking both together, if X and Y are both positive and absolute value of X is greater than absolute value of Y, X - Y can be 1/2 e.g. X = 2.5 and Y = 2
But if X and Y are both negative and absolute value of X is greater than absolute value of Y, X - Y will be negative. e.g. -2.5 -(-2) = -0.5.
Hence, if we combine the two statements, we are left with only X and Y both positive solution. Answer (C).

2. Is a(K) factor of b(M). K & M are exponents of a, b
1) A is a factor of b
2) K < M (Here K = M or K < M doesn't really matter)

Ques: Is $$b^M$$ divisible by $$a^K$$?
Let us just think for a moment here. When will $$b^M$$ be divisible by $$a^K$$. When a is a factor of b. In other words, when all the prime factors of a are in b as well so that they can get canceled e.g. 12 is divisible by which numbers? By 2, 3, 4, 6 etc i.e. numbers whose prime factors are in 12 as well and the power of these prime factors is less than or equal to that found in 12.
12 = 2^2 x 3. So a number having at the most two 2s and one 3 will divide 12. That is why 4 divides 12 but 9 doesn't.

Also K should be less than or equal to M to be certain that $$b^M$$ is divisible by $$a^K$$. Let me explain using an example: If a = 4 and b = 12, $$12^5$$ is divisible by $$4^5$$ but not by $$4^6$$ because 12 has one 4 and $$12^5$$ has five 4s so it is divisible by $$4^5$$. When I try to divide it by $$4^6$$, I need six 4s but I don't have them so it is not divisible. (Check theory of divisibility for further clarification.)
Hence both statements together are needed to solve the question. Answer (C).

In the third question, putting X = (A + B) and then checking if I can split this to be equal to F(A) + F(B) is pretty much what I would do. I would try the functions where terms are added or multiplied first because these are generally more symmetrical.
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Re: DS/PS Questions from GMAT software   [#permalink] 03 Nov 2010, 19:05
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