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(2) \(\frac{1}{k+1}> 0\)--> \(k+1>0\) --> \(k>-1\), hence \(k\) can be negative as well as positive: \(\frac{1}{k}\) may or may not be \(>0\). Not sufficient.
If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at random, what is the probability that their product will be in the form x^2 - (by)^2, where b is an integer?
(A) 1/2 (B) 1/3 (C) 1/4 (D) 1/5 (E) 1/6
Four expressions can be paired in \(C^2_4=6\) ways (choosing 2 out of 4);
\(x^2 - (by)^2=(x-by)(x+by)\) and only one pair is making such expression: \((x+1*y)(x-1*y)\), hence probability is 1/6.
In May Mrs Lee's earnings were 60 percent of the Lee family's total income. In June Mrs Lee earned 20 percent more than in May. If rest of the family's income did not change, then, in June, Mrs Lee's earnings were approximately what percent of the Lee family's total income ?
(A) 64% (B) 68% (C) 72% (D) 76% (E) 80%
Let total income of the family be 100$. In May Mrs Lee's earnings were 60 percent =60$. Rest of the family=40$. In June family=40$, Mrs Lee=60$*1.2=72$. Total=40+72=112$. Mrs Lees share=72/112=~64%.
(1) \(p=r\) --> \(\frac{1}{r}>\frac{r}{r^2+2}?\) --> as \(r^2+2\) is always positive, multiplying inequality by this expression we'll get:\(\frac{r^2+2}{r}>r?\) --> \(r+\frac{2}{r}>r?\) --> \(\frac{2}{r}>0?\). This inequality is true when \(r>0\) and not true when \(r<0\). Not sufficient.
(1) \(p=r\) --> \(\frac{1}{r}>\frac{r}{r^2+2}?\) --> as \(r^2+2\) is always positive, multiplying inequality by this expression we'll get:\(\frac{r^2+2}{r}>r?\) --> \(r+\frac{2}{r}>r?\) --> \(\frac{2}{r}>0?\). This inequality is true when \(r>0\) and not true when \(r<0\). Not sufficient.
(2) \(r>0\). Not sufficient by itself.
(1)+(2) \(r>0\), \(\frac{2}{r}>0\). Sufficient.
Answer: C.
Why we can't cross multiply?
r^2+2>r^2? _________________
Trying to make CR and RC my strong points
"If you want my advice, Peter," he said at last, "you've made a mistake already. By asking me. By asking anyone. Never ask people. Not about your work. Don't you know what you want? How can you stand it, not to know?" Ayn Rand
(1) \(p=r\) --> \(\frac{1}{r}>\frac{r}{r^2+2}?\) --> as \(r^2+2\) is always positive, multiplying inequality by this expression we'll get:\(\frac{r^2+2}{r}>r?\) --> \(r+\frac{2}{r}>r?\) --> \(\frac{2}{r}>0?\). This inequality is true when \(r>0\) and not true when \(r<0\). Not sufficient.
(2) \(r>0\). Not sufficient by itself.
(1)+(2) \(r>0\), \(\frac{2}{r}>0\). Sufficient.
Answer: C.
Why we can't cross multiply?
r^2+2>r^2?
Never ever multiply inequality by the expression with undefined sign. We can multiply both sides by r^2+2 because we know that this expression is always positive, BUT crossmultiplying involves multiplying by r, the sign of which is unknown. Hence we can not do that.
r^2+2>r^2 is wrong because of that. _________________
(1) \(p=r\) --> \(\frac{1}{r}>\frac{r}{r^2+2}?\) --> as \(r^2+2\) is always positive, multiplying inequality by this expression we'll get:\(\frac{r^2+2}{r}>r?\) --> \(r+\frac{2}{r}>r?\) --> \(\frac{2}{r}>0?\). This inequality is true when \(r>0\) and not true when \(r<0\). Not sufficient.
(2) \(r>0\). Not sufficient by itself.
(1)+(2) \(r>0\), \(\frac{2}{r}>0\). Sufficient.
Answer: C.
Why we can't cross multiply?
r^2+2>r^2?
Never ever multiply inequality by the expression with undefined sign. We can multiply both sides by r^2+2 because we know that this expression is always positive, BUT crossmultiplying involves multiplying by r, the sign of which is unknown. Hence we can not do that.
r^2+2>r^2 is wrong because of that.
Thanks! _________________
Trying to make CR and RC my strong points
"If you want my advice, Peter," he said at last, "you've made a mistake already. By asking me. By asking anyone. Never ask people. Not about your work. Don't you know what you want? How can you stand it, not to know?" Ayn Rand
The numbers x and y are not integers. The value of x is closest to which integer
(1) 4 is the integer that is closest to x+y. (2) 1 is the integer that is closest to x-y.
(1) 3.5<=x+y<4.5. Clearly insufficient to determine closest integer to x. Not sufficient. (2) 0.5<=x-y<1.5. Clearly insufficient to determine closest integer to x. Not sufficient.
(1)+(2) Adding inequalities from 1 and 2 --> 4<=2x<6 --> 2<=x<3 --> If 2<=x<2.5 closest integer 2 BUT if 2.5<=x<3 closest integer 3. Not sufficient.
Of the 200 members of a certain association, each member who speaks German also speaks English, and 70 of the members speak only Spanish. If no member speaks all three languages, how many of the members speak two of the 3 languages?
(1) 60 of the members speak only English (2) 20 of the members do not speak any of the three languages
Venn diagram is the best way to solve this question. So here are the tips for it: From the stem if 70 speaks only Spanish and no student speaks three languages, max # of students who speaks two languages is 200-70=130. Note that if ALL who speaks German speaks English means that no student speaks ONLY German, but not vise-versa, meaning that there may be the students who speak English only. Also note there may be the students among 200 who speak no English, German or Spanish.
So, basically we need to determine the number of students who speak only English and the number of students who doesn't speak any languages and subtract this from 130 (as we already subtracted only Spanish and know that there are no only German).
(1) 60 speaks ONLY English, max # of students who speaks two languages is 200-70-60=70. But we don't know how this 70 is split: don't know how many don't speak any of the languages. Not sufficient.
(2) 20 don't speak any of the language. Clearly insufficient.
(1)+(2) 70-20(any languages)=50 (# of students who speak two languages).
xy>0, is true if and only x and y have the same sign.
(1) x-y>-2, x and y can have the same sign or the opposite. Not sufficient. (2) x-2y<-6, or -x+2y>6, x and y can have the same sign or the opposite. Not sufficient.
(1)+(2) Adding inequalities we'll get y>4 and x>2. So x and y are both positive. Sufficient.
xy>0, is true if and only x and y have the same sign.
(1) x-y>-2, x and y can have the same sign or the opposite. Not sufficient. (2) x-2y<-6, or -x+2y>6, x and y can have the same sign or the opposite. Not sufficient.
(1)+(2) Adding inequalities we'll get y>4 and x>2. So x and y are both positive. Sufficient.
Answer: C.
can we add inequilaties ?
as (1) - (2) ==> y >-2-(-6) ==> y>4 ?
can we do this ? _________________
Thanks, Sri ------------------------------- keep uppp...ing the tempo...
Press +1 Kudos, if you think my post gave u a tiny tip
xy>0, is true if and only x and y have the same sign.
(1) x-y>-2, x and y can have the same sign or the opposite. Not sufficient. (2) x-2y<-6, or -x+2y>6, x and y can have the same sign or the opposite. Not sufficient.
(1)+(2) Adding inequalities we'll get y>4 and x>2. So x and y are both positive. Sufficient.
Answer: C.
can we add inequilaties ?
as (1) - (2) ==> y >-2-(-6) ==> y>4 ?
can we do this ?
You can add inequalities ONLY when their signs are in the same direction: a>b, c>d --> a+c>b+d.
OR You can subtract inequalities ONLY when their signs are in the opposite directions: a>b, d<c a-d>b-c
So in our case: x-y>-2 -x+2y>6 Adding them: y>4
OR x-y>-2 x-2y<-6 Subtract them: y>4 the same thing. _________________
+ 1 kuduos to Bunuel from me ,, as i have been following your tips which helped me solve all these question correctly . could nt have done with out u.. keep u the good work .
Srini, if you dont mind me asking, how did you end up doing on the quant section for preptest 2?
GMATPrep2 640 (Q 23/37, V 28/41) taken on 11/01/2009 (forgot to note scaled scores-sorry) GMATPrep1 650 (Q49, V29; 26/37, 28/41) taken on 10/25/2009 _________________
Thanks, Sri ------------------------------- keep uppp...ing the tempo...
Press +1 Kudos, if you think my post gave u a tiny tip
Thanks for your reply. The reason I asked was because I was curious to know the level of difficulty I would expect according to the Preptest 2 questions that you have provided. Was this a score of 37?