BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

A bunch of difficult DS questions - DS- 1 to 4

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 49
Joined: Sun Jan 13, 2008 10:47 pm
Thanked: 1 times

A bunch of difficult DS questions - DS- 1 to 4

by desihokie » Tue Jan 15, 2008 12:11 am
Atleast I find them difficult!! I get extremely confused when it comes to inequalities in DS questions. Any strategy?

Edit: I cut all the questions but 1-4 which have been answered in this post. Again, thanks everyone who helped me.
Attachments
DS_1-4.jpg
Last edited by desihokie on Tue Jan 15, 2008 8:54 am, edited 2 times in total.
Top
Source: — Data Sufficiency |
Legendary Member
Posts: 645
Joined: Wed Sep 05, 2007 4:37 am
Location: India
Thanked: 34 times
Followed by:5 members

by camitava » Tue Jan 15, 2008 1:36 am
desihokie, refer my answers below -

1. C
2. A
3. E
4. A
5. C
6. C
7. C
8. D
9. C
10. E
11. D
12. B

Can you pls check the answers and let us know whether they are fine or not!
Correct me If I am wrong


Regards,

Amitava
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 423
Joined: Thu Dec 27, 2007 1:29 am
Location: Hyderabad, India
Thanked: 36 times
Followed by:2 members
GMAT Score:770

by simplyjat » Tue Jan 15, 2008 2:35 am
It would have been better if you listed each question in a separate thread. Anyways...

Question 1. Answer C

(-a, b) and (-b, a) signifies that both a and b have the same sign; either both positive or both negative. if a & b had different signs, (-a, b) and (-b,a) would be in separate quadrants.

When a,b are both positives, (-a, b) and (-b, a) both lie in second quadrant, and a,b are both negatives (-a, b) and (-b, a) both lie in the fourth quadrant.

So in order to check (-x, y) region, we need to know two things.
1. whether x & y have the same sign or not.
2. whether x,y & a,b have the same sign or not.

The first information tells us that both x & y have the same sign. And the second information tells us that both x * a have the same sign. Thus all of the variable have the same sign and consequently reside in one quadrant (Either first or third)
Last edited by simplyjat on Tue Jan 15, 2008 2:44 am, edited 1 time in total.
simplyjat
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 423
Joined: Thu Dec 27, 2007 1:29 am
Location: Hyderabad, India
Thanked: 36 times
Followed by:2 members
GMAT Score:770

by simplyjat » Tue Jan 15, 2008 2:43 am
Question 2, Answer C

There is a list and there are some elements in the list. The product of all the elements in the list is positive if
1. All the elements of the list are positive.
2. There are even number of negative elements in the list.

The first information tells us that either all the numbers are positive or all the numbers are negative. So if there are odd number of elements in a list comprised of all negative numbers, the result will be negative. So this alone is insufficient.

The second information tells us that there are even number of elements in the list, but does not tell anything about the elements and thus insufficient.

But if we combine the two informations, we have a list of either all positive or all negative numbers, containing even elements. And in both the cases the product of all the elements will be negative.
simplyjat
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 423
Joined: Thu Dec 27, 2007 1:29 am
Location: Hyderabad, India
Thanked: 36 times
Followed by:2 members
GMAT Score:770

by simplyjat » Tue Jan 15, 2008 2:59 am
Question 3 Answer A.

Let x be the total number of marbles (x = r+w+b) , then probability of getting a red marble is r/x, and probability of white marble is w/x.

Basically the question is asking whether r > w or not...

First information tells us that r / (b+w) > w / (b+r)
=> r / (x-r) > w / (x - w) [ because x = r+w+b ]
=> r * (x -w) > w * (x-r) [ we can do the multiplication as all the number are positve]
=> rx -rw > wx -rw
=> rx > wx [ subtracting rw from each side ]
=> r > w [dividing x from each side as x is positive]
so the probability of red is more than probability of white marble.

For the second information I am not sure whether b-w > r or b = w > r.
If it is b-w > r then the information is insufficient and the answer is B,
If it is b=w > r then we have a contradiction with first statement.
Last edited by simplyjat on Tue Jan 15, 2008 3:19 am, edited 1 time in total.
simplyjat
Top
Master | Next Rank: 500 Posts
Posts: 195
Joined: Sun Oct 21, 2007 4:33 am
Thanked: 10 times

by sankruth » Tue Jan 15, 2008 3:01 am
Q 3

A

St 1
If ratio of Red : Rest of the balls is greater than White : Rest of the balls then R > W So SUFF

St2
b-w > r says nothing about r and w hence INSUFF
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 423
Joined: Thu Dec 27, 2007 1:29 am
Location: Hyderabad, India
Thanked: 36 times
Followed by:2 members
GMAT Score:770

by simplyjat » Tue Jan 15, 2008 3:16 am
Question 4 Answer C

According to the question we can represent n & t as follows
n = 3x + 2 and t = 5y + 3, where x and y are any arbitrary numbers.
so nt = (3x + 2) * (5y + 3)
= 15xy + 9x + 10y + 6
And we have to find out the remainder when the above expression is divided by 15.
The first term of the expression is a multiple 15 and thus will always give remainder zero and thus we can reduce the equation to
9x + 10y + 6

Now the first information tells us the n - 2 is divisible by 5, i.e. (3x+2) - 2 => 3x is completely divisible by 5. It can be inferred that 3 (3x) will be divisible by 3 (5) =15. So in the equation we can remove 9x and are left with 10y + 6, so this information is insufficient.

From the second information we know that t is divisible by 3, i.e. (5y + 3) is divisible by 3. thus 5y is divisible by 3 so 10y will also be divisible by 3. And 10y is already divisible by 5. So 10y is divisible by 5*3 = 15. so we can remove 10y from the expression. and are left with 9x + 6. so this information alone is insuffificent.

Now if we use both the information we can remove both 9x and 10y from the equations and left with 6 which is the remainder when nt is divided by 15
Last edited by simplyjat on Tue Jan 15, 2008 7:51 am, edited 1 time in total.
simplyjat
Top
Master | Next Rank: 500 Posts
Posts: 195
Joined: Sun Oct 21, 2007 4:33 am
Thanked: 10 times

by sankruth » Tue Jan 15, 2008 3:28 am
simplyjat wrote:Now if we use both the information we can remove both 9x and 10 from the equations and left with 6 which is the remainder when nt is divided by 15
I used a cumbersome method of substituting numbers to arrive at the same answer. However I like your method as it is short and can be done inside 2 minutes. Can you please elaborate tha last bit about using both information. i.e. How did you eliminate 9x and 10y
Top
Master | Next Rank: 500 Posts
Posts: 195
Joined: Sun Oct 21, 2007 4:33 am
Thanked: 10 times

by sankruth » Tue Jan 15, 2008 3:42 am
sankruth wrote:
simplyjat wrote:Now if we use both the information we can remove both 9x and 10 from the equations and left with 6 which is the remainder when nt is divided by 15
I used a cumbersome method of substituting numbers to arrive at the same answer. However I like your method as it is short and can be done inside 2 minutes. Can you please elaborate tha last bit about using both information. i.e. How did you eliminate 9x and 10y
I figured it myself. Was a true Homer Simpson moment a few minutes ago... :lol: Hope I dont any these moments during the test :wink:
Top
User avatar
Legendary Member
Posts: 986
Joined: Wed Dec 20, 2006 11:07 am
Location: India
Thanked: 51 times
Followed by:1 members

by gabriel » Tue Jan 15, 2008 6:08 am
@desihokie (Lol at the name btw) .. please follow forum rules, just one question per thread.

Regards
Top
Senior | Next Rank: 100 Posts
Posts: 49
Joined: Sun Jan 13, 2008 10:47 pm
Thanked: 1 times

by desihokie » Tue Jan 15, 2008 8:56 am
Thanks everyone for your help.

Mod: I moved all but Q-1-4 which have already been answered here. Will post the rest seperately.

-desihokie (a desi from VT)!
Top
Master | Next Rank: 500 Posts
Posts: 128
Joined: Sat Nov 17, 2007 11:59 am
Thanked: 1 times

by cris » Tue Jan 29, 2008 5:51 pm
Anybody knows the correct answers to these questions?

What about question 4, any of you has found a more simple way of attacking "remainder" problems??

Thanks!
Top
Post new topic Post Reply
• Page 1 of 1