tricky DS questions

This topic has expert replies
Newbie | Next Rank: 10 Posts
Posts: 8
Joined: Wed Jan 12, 2011 6:00 am

tricky DS questions

by akane » Sun Oct 09, 2011 8:26 am
Hello!
Can anyone help me understand these two DS questions and how you get to the solution? thx

#1: lines n and p lie in the xy plane. Is the slope of line n less than the slope of line p?
(1) lines n and p intersect at point (5,1)
(2) the y-intercept of n is greater that the y-intercept of p.

#2: On a certain sight-seeing tour, the ratio of the number of women to the number of children was 5 to 2. What was the number of men on the tour?
(1) on the tour, the ratio of the number of children to the number of men is 5 to 11
(2) the number of women is less than 30

For each of these, I worked my way up to combining 1and2 so I knew the answer was either C or E but could not figure it out. both answers are C

Master | Next Rank: 500 Posts
Posts: 197
Joined: Thu Sep 15, 2011 10:22 am
Thanked: 6 times
Followed by:2 members

by parul9 » Sun Oct 09, 2011 8:51 am
I think qsn 1 is C.
1 Alone cant tell the slope
2 Alone could hold true for parallel lines too.

If we combine both, then we have two lines intersecting t a point with the y intercept of one greater than the other. So the slope comparison can be done.

For qsn 2, I got E as the answer.
w = no. of women
c = no. of children
m = no. of men

given: w/c = 5/2.

1. c/m = 5/11
We cant get the "number" of men based on ratios alone.
So, insufficient.

2. w<30.
But W could be any number below 30 (and perhaps a multiple of 5, considering the ratio of w/c)
So, this too is insufficient.

If we combine 1 and 2, we have c/m = 5/11
and w<30, and given info that w/c = 5/2.
I dont think we can get an answer from this either. So I think I'd have marked E for this.

But since you already said the answer is C for both, I am trying to C-ify my though process! :D
Since c/m = 5/11, so, c and m can be taken as 5n and 11n. So, c should be a multiple of 5.
Now, w/c = 5/2, so w and c can be taken as 5m and 2m. But since c is a multiple of 5, 2m should also be a multiple of 5.
since w<30, m should be 5, as w/c = 25/10 is the only ratio that can satisfy the criterion of c being a multiple of 5.
so, if c = 10, then using ratio c/m = 5/11, we can get m=22.
So perhaps C is indeed the answer!

I am not sure if my reasoning for C is correct!
I would have anyway marked the answer as E had I not known this beforehand.
:(

Senior | Next Rank: 100 Posts
Posts: 61
Joined: Tue Apr 13, 2010 3:29 am
Thanked: 8 times

by n@resh » Sun Oct 09, 2011 9:21 am
#1: lines n and p lie in the xy plane. Is the slope of line n less than the slope of line p?
(1) lines n and p intersect at point (5,1)
(2) the y-intercept of n is greater that the y-intercept of p.

Ans: C!

From stat(1): we have a point of intersection:(5,1) for the lines n ( Yn = -(Bn/An)Xn + Bn )
And line p: ( Yp = -(Bp/Ap)Xp + Bp ) . Here we also know slopes are -(Bn/An) and -(Bp/Ap) for the lines n and p respectively. So, we need to find is -(Bn/An) < -(Bp/Ap)?
we have: 1 = -(Bn/An)5 + Bn --- I and 1 = -(Bp/Ap)5 + Bp --- II
from I and II, slopes of n and p's are ( 1 - Bn )/5 and ( 1 - Bp )/5
it's still not sufficient to compare those above values.

From stat(2): Y intercept of n > Y intercept of p
consider Yn = -(Bn/An)Xn + Bn and Yp = -(Bp/Ap)Xp + Bp..
i.e. Bn > Bp.
But, it still not sufficient compare slopes.

From using both statments: now we have ( 1 - Bn )/5 and ( 1 - Bp )/5 and Bn > Bp

so we can conclude: values of ( 1 - Bn )/5 < ( 1 - Bp )/5 Hence slope of line n < slope of line p.

#2: On a certain sight-seeing tour, the ratio of the number of women to the number of children was 5 to 2. What was the number of men on the tour?
(1) on the tour, the ratio of the number of children to the number of men is 5 to 11
(2) the number of women is less than 30

Ans:
From stat(1): C/M = 5/11 and given W/C = 5/2 , asking for number of men (M) = ?
=> C = (5/11)M = (2/5)W i.e. M = ( 22/25) W ?!
Just not sufficient!

From stat(2): Women < 30!
clearly, not possible as no information on Men.

from using both statments: M = ( 22/25) W and W < 30.
now we can see there is only one possible value of W i.e. 25,
Hence no.of Men = 22.

Answer:C!

User avatar
GMAT Instructor
Posts: 349
Joined: Wed Sep 28, 2011 3:38 pm
Location: Austin, TX
Thanked: 236 times
Followed by:54 members
GMAT Score:770

by GmatMathPro » Sun Oct 09, 2011 10:03 am
akane wrote:Hello!
Can anyone help me understand these two DS questions and how you get to the solution? thx

#1: lines n and p lie in the xy plane. Is the slope of line n less than the slope of line p?
(1) lines n and p intersect at point (5,1)
(2) the y-intercept of n is greater that the y-intercept of p.
Statement 1: Picture two lines on the graph that go through point (5,1). We can rotate these two lines around this point any way we want without violating the conditions. This would change the slopes, and we could certainly rotate them in a way that gives multiple answers to the question. INSUFFICIENT.

Statement 2: Basically the same idea as statement 1. This just establishes a point on each of the lines, but each line could still have any possible slope.


Statement 1&2 together: Now, both points go through (5,1), and one line has a greater y-intercept than the other, so one HAS to be steeper than the other. there's no point in worrying about which one it is. we just know that we could definitely answer the question.

Ans: C
#2: On a certain sight-seeing tour, the ratio of the number of women to the number of children was 5 to 2. What was the number of men on the tour?
(1) on the tour, the ratio of the number of children to the number of men is 5 to 11
(2) the number of women is less than 30
Given: W:C=5:2

Statement 1: C:M=5:11. Multiply W:C=5:2 by 5 to make it W:C=25:10. Multiply C:M=5:11 by 2 to make it C:M=10:22. Now we can write a three-way ratio as W:C:M=25:10:22. This establishes the basic ratio, so we know any distribution of women, children, and men could be exactly 25,10, and 22, or it could also be any multiple of that ratio: (50,20,44);(75,30,66)......INSUFFICIENT.

Statement 2: Knowing that the number of women is less than 30 tells us nothing about the number of men. INSUFFICIENT.

Statements 1&2: From statement 1 we know the number of women, children and men(W,C,M) must be (25,10,22) or some integer multiple: (25,10,22);(50,20,44);(75,30,66);(100,40,88)....and so on. However, only the first one satisfies statement 2 which says that the number of women is less than 30, meaning the number of men must be exactly 22. SUFFICIENT.

Ans: C
Pete Ackley
GMAT Math Pro
Free Online Tutoring Trial

Newbie | Next Rank: 10 Posts
Posts: 8
Joined: Wed Jan 12, 2011 6:00 am

by akane » Sun Oct 09, 2011 12:24 pm
Thanks to all of you who took the time to try out this problems and provide your results. it was very helpful. I get it now.
@parul9: I went with E on the 2nd question too, but your reasoning for getting to C is right, it has to do with multiple of 5 etc... GmatMathPro went trhu it very well. take a look @ what he wrote if you are curious to see how your reasoning and his post compare.

@n@resh: Thx for your post. I didn't understand why the slopes in the equations of the lines are negative. ex: in this equation Yp=-(Bp/Ap)Xp+Bp.

@GmatMathPro: Thx for your detail explanation. I could only get all the way to C or E with both questions using the same methods you described in your post but got stuck. really appreciate it :) !

Newbie | Next Rank: 10 Posts
Posts: 4
Joined: Wed Jul 13, 2011 11:09 am

by battlefield » Sun Oct 09, 2011 3:17 pm
Hi Pete,
I really appreciate your efforts in solving so many problems on BTG.

I have a question on the understanding of the y intercept.
What is meant by a greater intercept?

Isn't it that an intercept of -6 is greater then an intercept of 3?
-6 is greater in number than 3 but its just on the opposite side of the graph.

Also, is a negative slope of -1/2 lesser than a positive slope of 1/3 ?

User avatar
GMAT Instructor
Posts: 349
Joined: Wed Sep 28, 2011 3:38 pm
Location: Austin, TX
Thanked: 236 times
Followed by:54 members
GMAT Score:770

by GmatMathPro » Sun Oct 09, 2011 3:51 pm
battlefield wrote:Hi Pete,
I really appreciate your efforts in solving so many problems on BTG.

I have a question on the understanding of the y intercept.
What is meant by a greater intercept?

Isn't it that an intercept of -6 is greater then an intercept of 3?
-6 is greater in number than 3 but its just on the opposite side of the graph.

Also, is a negative slope of -1/2 lesser than a positive slope of 1/3 ?
Say line n is represented by the equation y=ax+b and line p is represented by the equation y=cx+d. If line n has a greater y-intercept than line p, then b>d. Graphically, the place where line n intersects the y-axis would be higher than the place where line p intersects the y-axis.

-6 is not a greater intercept than an intercept of 3. the exact same rules apply as with simple inequalities. -6 is not greater than 3, so -6 is not a greater intercept than 3. If the wording were something like, "the MAGNITUDE of the y-intercept of line n is greater than the MAGNITUDE of the y-intercept of line p", then you would have a point because the word magnitude implies the concept of absolute size or absolute value, and -6 has a bigger magnitude than 3. But if it's just simply "c is greater than d", treat it the same as you would treat c>d

Same thing with slopes. A negative slope is always less than a positive slope no matter what. So yes, a slope of -1/2 is less than a slope of 1/3. If you picture a line on a graph as something you have to walk on from left to right, the bigger the slope gets, the harder it is to walk up it from left to right. The smaller it gets, the easier it is to walk up it from left to right. When it becomes zero, it's like walking on flat ground. When it's negative, it's like walking downhill. The smaller(i.e. more negative) it gets, the steeper the downhill slope is from left to right.

Does that help?
Pete Ackley
GMAT Math Pro
Free Online Tutoring Trial

Newbie | Next Rank: 10 Posts
Posts: 4
Joined: Wed Jul 13, 2011 11:09 am

by battlefield » Sun Oct 09, 2011 9:52 pm
That's really really helpful Pete. I highly appreciate it.

Senior | Next Rank: 100 Posts
Posts: 55
Joined: Tue Jul 27, 2010 8:37 pm
Thanked: 11 times

by KrazyKarl » Mon Oct 10, 2011 9:55 pm
For the first question, it's hard to explain without drawing but I drew lines to see if I could make the one with the higher y-intercept have a higher slope and I couldn't do it. Because they both end up in the upper right quadrant (is that #1? I should remember that) then they could either both be sloping up and to the right (both positive slope), both down and to the right (both negative) or one of each. But every time I drew it the one that started higher on the y-intercept had a smaller slope (either steeper both negative, or less steep both positive, or had to be the negative one with a +/-). So by drawing and trying to get different answers I figured out that I couldn't so C had to be sufficient.