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Manager
Joined: 08 Jul 2006
Posts: 89

Artemov wrote: The differential distance of 2 km is being covered in 1hr. So, 0.2 km would be covered in 0.1 hr. And the deer will travel 1 km in 0.1 hr @ 10km/hr.
I would choose your solution if you would explain further.
I understand the differential distance being covered in 1hr, I understand that 0.2km will be covered in 0.1hr.
Notwithstanding the fact that you got the right answer, I don't understand why we are using 0.1hr to arrive at the distance traveled by the deer before it is caught when, clearly, the lion took that time to arrive at the deer's initial position. Shouldn't the time we use be greater?



Senior Manager
Joined: 05 Oct 2006
Posts: 257

i tried this way.
when taking both choices i.e., x>2y and x<y+2
we can write, 2y<x<y+2
so x lies in between 2y and y+2 for all x,y>0
from 2y<x<y+2
implies 2y<y+2
impliws y<2
hence we can answer the question that y is not more than 2.
hencechoice c.
any comments plz??
ywilfred wrote: If x and y are positive, is y > 2? 1. x > 2y 2. x < y + 2
From statement (1), we have x > 2y => y < x/2. So y can take on various positive values dependign on what postive values x takes on. Statement (1) is therefore not sufficient.
From statement (2), we have x < y + 2 => x  y < 2. So now x and y can take on sets of value, limited by the inequality above. If x,y = 3,2, then x  y < 2 but y is not greater than 2. However, if x,y = 10,9, then x y < 2 but y is now greater than 2. So statement (2) insufficient.
Using both statement (1) and statement (2), we have y < x/2 and xy < 2. Now, we'll test for possible x,y pairs that satisfies both inequalities.
If x = 10, y < 5. (say, y = 4), then xy = 104 > 2. So x cannot be any value larger than 10.
If x = 5, y < 2.5 (say y = 2.4), then xy = 2.6 > 2. So x cannot be any value larger than 5.
If x = 3, y < 1.5 (say y = 1.4), then xy = 1.6 < 2. So x can be 3 and below.
If x = 4, y < 2 (say y = 1.9), then xy = 2.1 > 2. So x cannot be 4 and above.
If x = 3.59, y < 1.795 (say y = 1.69) then xy = 1.9 < 2. So Maximum value of x would be approximately 3.59 and y would be less than 2 for all values of x 3.59 and below.
The answer is therefore C.
Note: The question does not say x and y are integers, so we must also consider decimals (or fractions)



Manager
Status: Post MBA, working in the area of Development Finance
Joined: 09 Oct 2006
Posts: 154
Location: Africa

Rayn wrote: Artemov wrote: The differential distance of 2 km is being covered in 1hr. So, 0.2 km would be covered in 0.1 hr. And the deer will travel 1 km in 0.1 hr @ 10km/hr. I would choose your solution if you would explain further.I understand the differential distance being covered in 1hr, I understand that 0.2km will be covered in 0.1hr. Notwithstanding the fact that you got the right answer, I don't understand why we are using 0.1hr to arrive at the distance traveled by the deer before it is caught when, clearly, the lion took that time to arrive at the deer's initial position. Shouldn't the time we use be greater?
A leopard spots a deer from a distance of 200 meters. As the leopard starts chasing the deer, the deer also starts running. Given that the speed of the deer is 10 km/h and that of the leopard is 12 km/h, how far would have the deer run before it is caught?
How much distance would the deer cover in 0.1 hr => 1 km
How much distance would the leopard cover in 0.1 hr => 1.2 km
At the end of 0.1 hr the lead of 200 mtrs would have been covered.



Director
Joined: 26 Feb 2006
Posts: 854

ywilfred wrote: Is sqrt(x5)^2 = 5x?
1) xx > 0 2) 5x > 0
sqrt(x5)^2 is really just (x5). So what the question is asking is whether x5 = 5x => x = 10. If either statement tells us if x is 10 or any other particular value, then it is sufficient.
From statement (1), for xx to be greater than zero, x must be negative. And if x is negative, then it can't be 10. Therefore, statement (1) is sufficient.
From statement (2), we have 5  x > 0 => x < 5. We know x has to be less than 5, so x cannot be 10. Statement (2) alone is sufficient as well.
The answer is therefore D.
i do this way.
the question is "Is sqrt(x5)^2 = 5x?" or "is x = 5?".
from 1) we know x is ve. so x is not equal to 5. suff.
from 2) 5>x. so x is not equal to 5. suff.
so D......



Intern
Joined: 23 Jul 2005
Posts: 7

ywilfred wrote: If x and y are nonzero integers, is x ^y < y ^x
1. x = y^2 2. y > 2
For such questions, the statement is sufficient if we can answer a yes or a no to the questions "Is x^y < y^x".
Using statement (1), we have x = y^2. So if y = 1, x = 1. then x^y is not less than y^x. However, if y = 3, x = 9. Then x^y is less than x^y. So we can't answer the question posed and statement (1) is therefore not sufficient.
Using statement (2), we have y > 2, but we have no values of x to work with. So we still can't answer the question. Statement (2) is therefore not sufficient.
Using both statement (1) and statement (2), we know y is 3 and above. For all values of y that is 3 and above, y^x is always less then x^y. So we can now answer the question  "Is x^y < y^x ?" The answer is no.
The answer therefore is C.
I agree with the answer C but disagree with the last statement  in bold. Indeed x^y < y^x



Intern
Joined: 18 Jan 2007
Posts: 2
Location: India

ywilfred wrote: What is the least value of the three digit integer y?
1) the sum of the three digits is 5 2) y is divisible by 5
From statement (1), we're told the sum of the three digits is 5. The least value will therefore be in the range of 100+. 104 should therefore be the smallest number. Statement (1) is sufficient.
From statement (2), were told y is divisible by 5. So the last digit should be either 0 or 5. So the number can either be 100 or 105. Since we're asked for the least value, therefore y has to be 100. Statement (2) is therefore sufficient.
The answer should be D.
For Statement (1) d least value should be 995 as it isnt mentioned dat the 3 digit number is positive.
For Statement (2) The least value is 100.
What answer choice should we choose in this case? E? or is it D?



Intern
Joined: 23 Jul 2005
Posts: 7

dipidha wrote: ywilfred wrote: What is the least value of the three digit integer y?
1) the sum of the three digits is 5 2) y is divisible by 5
From statement (1), we're told the sum of the three digits is 5. The least value will therefore be in the range of 100+. 104 should therefore be the smallest number. Statement (1) is sufficient.
From statement (2), were told y is divisible by 5. So the last digit should be either 0 or 5. So the number can either be 100 or 105. Since we're asked for the least value, therefore y has to be 100. Statement (2) is therefore sufficient.
The answer should be D. For Statement (1) d least value should be 995 as it isnt mentioned dat the 3 digit number is positive. For Statement (2) The least value is 100. What answer choice should we choose in this case? E? or is it D?
well that is a good point. How can you associate the ve sign to first 9 and positive sign to second 9. I dont agree with that. May be 500 is the lowest 3 digit number which adds to 5.
Also for second statement may be 995 is the right one.
However the answer still remains as D



SVP
Joined: 03 Jan 2005
Posts: 2108

Good observations. I would say that this question is not well designed. It may be better if it had said "positive integers". The other thing though, is that from my experience, for GMAT questions that have D as the answer, the two options should lead to the same result.
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Intern
Joined: 18 Jan 2007
Posts: 2
Location: India

thats exactly my doubt! how do v say its D wen d 2 ans differ??



Senior Manager
Joined: 09 Aug 2006
Posts: 498

ywilfred wrote: I'll start with this one today.
A slot machine in a Las Vegas casino has an average profit of $600 for each 8 hours shift for the five days sunday through thrusday , inclusive . If the average per shift profit on Friday and saturday is 25% greater than on the other days of the week and the slot machine is in operation every hour of everyday, wheat is the total weekly profit that the casino makes fromt he slot machine? $4500 $9000 $13500 $15500 $27000
This questions tests on your ability to breakdown the long passage, assimilate the information and turn them into equations to work on.
We're told the slot machines has:
1. Average profit/shift = $600 for Sunday  Thursday 2. Average profit/shift = 25% greater on Friday and Saturday 3. Slot machine is in operation every hour of everyday
Since we know the machine works non stop for 24 hours each day, and there's a 8 hour shift, so each day there's 24/8 = 3 shifts So, from Sunday to Thursday, there's a total of 5*3 = 15 shifts So the total profit from the 5 days = 600*15 = $9000
We also know from the question, that the average profit/shift = (125/100)*600 = $750 And for the two days, we have 2*3 = 6 shifts Therefore, the total profit for Friday and Saturday = 750*6 = $4500
So all we need to do now is to add up the two total profits to give the total weekly profit. That sum is $9000+$4500 = $13,500. The answer is therefore, C.
Hi ywilfred,
I have a doubt.
We have already calculated the profit for 5 days as 9000$ including the thursaday and Fri . This 9000$ profit includes the 600$/shift profit on thursday and Frid. Since thur and fri have 25% more profit /shift which is 750$/shift x 6 = 4500$. But out of this 4500$ 3600$ have already been taken into account. So I think the effective profit is $9000 + (4500  3600) = 9000 + 900 = 9900$. Adding 4500 means we are adding the profit twice. Even though there is no option in answer choice, but my reasoning says this. Could you please suggest where am I wrong.



Director
Joined: 14 Sep 2007
Posts: 898
Schools: Kellogg '10

Not sure if anyone will find this helpful, but I've pasted all of the questions from this thread into a single Word document.
This will be my airplane study guide.
Thanks so much to ywilfred for putting this thread together.



Senior Manager
Joined: 27 Aug 2007
Posts: 250

Artemov wrote: ywilfred wrote: A leopard spots a deer from a distance of 200 meters. As the leopard starts chasing the deer, the deer also starts running. Given that the speed of the deer is 10 km/h and that of the leopard is 12 km/h, how far would have the deer run before it is caught?
A. 3 km B. 4 km C. 2 km D. 1 km E. 5 km
This question test the ability to recall the DistanceSpeedTime formula, and also to manipulate the question to formulate the equation for solving the question (by use of algebra). In such DistanceSpeedTime question (which involves A catching up with B), there's normally one quantity that's similar. Also, remember to make sure that your units match, otherwise your answer will not come out right.
We know from the passage: 1. The Leopard and the Deer are 200 meters apart = 0.2 km 2. THe speed of the Leopard = 12km/hr 3. The speed of the Deer = 10 km/hr
Let's assume that the Deer has travelled x km before it is caught. The Leopard would then have to travel (x+0.2)km. There is one quantity that is the same here, that is the time taken for the deer to travel that x km, and for the Leopard to travel (x+0.2km).
Since we know that Time = Distance/Speed, we can equate:
Time taken by Deer to travel x km = Time taken by Leopard to travel (x+0.2) km x/10 = (x+0.2)/12 12x = 10x + 2 2x = 2 x = 1 km
The answer is therefore, D.
The differential distance of 2 km is being covered in 1hr. So, 0.2 km would be covered in 0.1 hr. And the deer will travel 1 km in 0.1 hr @ 10km/hr.
Have the same as Artemov's:
10t+0.2=12t
t=0.1
0.1*10=1 km



Manager
Joined: 07 Apr 2009
Posts: 140

Re: Beginners Forum (Reworked questions)
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08 Jul 2009, 10:36
Quote: If X and Y are integers and Y =/x+3/ + /4x/, does y equal 7?
1. x < 4 2. x > 3
Using statement (1), we know x can be 3 and below. This will give us a number of y values, so statement (1) is not sufficient.
Using statement (2), the same problem exists. x can be any value above 3 and this will give us a number of y values. So statement (2) is not sufficient.
Using both statements, we can get the inequality: 3 < x < 4.
Now x will have to take on either values 2, 1, 0, 1, 2 or 3.
If x = 0, y = 7. If x = 1, y = 7 If x = 2, y = 7 If x = 3, y = 7 If x = 1, y = 7 If x= 2, y = 7
So using both statements will limit values of x that always result in y taking on the value of 7.
The answer is therefore C.
why isnt the answer B ? For all values I get only 7 , I must be overlooking something, help !



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Joined: 14 Feb 2010
Posts: 46
Location: Tokyo
Schools: Insead

Rubashov1 wrote: Not sure if anyone will find this helpful, but I've pasted all of the questions from this thread into a single Word document.
This will be my airplane study guide.
Thanks so much to ywilfred for putting this thread together. nice! very helpful!



Intern
Joined: 01 May 2010
Posts: 4

Himalayan wrote: ywilfred wrote: Is sqrt(x5)^2 = 5x?
1) xx > 0 2) 5x > 0
sqrt(x5)^2 is really just (x5). So what the question is asking is whether x5 = 5x => x = 10. If either statement tells us if x is 10 or any other particular value, then it is sufficient.
From statement (1), for xx to be greater than zero, x must be negative. And if x is negative, then it can't be 10. Therefore, statement (1) is sufficient.
From statement (2), we have 5  x > 0 => x < 5. We know x has to be less than 5, so x cannot be 10. Statement (2) alone is sufficient as well.
The answer is therefore D. i do this way. the question is "Is sqrt(x5)^2 = 5x?" or "is x = 5?". from 1) we know x is ve. so x is not equal to 5. suff. from 2) 5>x. so x is not equal to 5. suff. so D...... I agree with the second point of view of Himalayan. I can't really understand how the first one is correct. x5 = 5x => x = 10. < how on earth is this right??? quite helpful post nevertheless there are some mistakes though...



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Re: Beginners Forum (Reworked questions)
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01 Feb 2012, 10:13
hi , just would like to know as to how many problems are there in this section? as of now i have seen 19 questions in this page 1. request to let me know thanks eshwar
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regards eshwar



Manager
Joined: 06 Jan 2012
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Scrat wrote: Rubashov1 wrote: Not sure if anyone will find this helpful, but I've pasted all of the questions from this thread into a single Word document.
This will be my airplane study guide.
Thanks so much to ywilfred for putting this thread together. nice! very helpful! +1, Thanks for your efforts!



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Re: Beginners Forum (Reworked questions)
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31 Jan 2018, 01:09
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Re: Beginners Forum (Reworked questions) &nbs
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