Data Sufficiency

I am struggling to find answers for the GMATPrep questions on BTG and anywhere else. My search returns all kinds of stuff but not the answers for the questions I am looking for. If there is a book I can buy I am more than happy to get all the answers, until then it would be great to see how to solve these. Thanks so much.

answer to 1st:
A: highest= g, and smallest = g-r (as r is the range)
B: h, h-s
g-r > h-s?? or g+s> h+r??
A) r<s
Insufficient. no info about g,h
B) g>h
Insufficient. no info about r and s.

A&B) s>r and g>h
thus g+s > h+r

IMO C

answer to 2nd:
x=-a and -b.
A)a+b = -1. insufficient
(a=b=-0.5 possible and also a=-0.7 and b=-0.3 possible)
B) (0,-6) thus ab = -6 insufficient

A&B) a+b = -1 and ab=-6
-3,2 are the only 2 values.
Sufficient
IMO C

answer to 3rd:
m/r =x/y ????
A) m/y = x/r or mr=xy. Insufficient
B) m+x / r+y = x/y
so we have to check whether (m+x)/(r+y) = m/r
insufficient
x,y are different..

A&B) x=mr/y
thus (m+x)/(r+y) = (m+ (mr/y))/(r+y) = m (r+y) / y(r+y) = m/y
true only when y=r. insufficient

IMO E

4th:
y miles: x m/h and 40-y: 1.25x
t1 = y/x + (40-y)/1.25x = 160-y / 5x
t2 = x m/h for 40 miles. means t2 = 40/x hr.
t1 is what percent of t2?? let p%
then t1 = p/100 * t2 -> (160-y)/5x = p/100 * 40/x
p = (160-y)/2

B) gives value of y. thus sufficient
IMO B

Thank you so much for all the amazing help. This one is a bit unclear since GMATPrep software says B is the right answer.

cans wrote:answer to 3rd:
m/r =x/y ????
A) m/y = x/r or mr=xy. Insufficient
B) m+x / r+y = x/y
so we have to check whether (m+x)/(r+y) = m/r
insufficient
x,y are different..

A&B) x=mr/y
thus (m+x)/(r+y) = (m+ (mr/y))/(r+y) = m (r+y) / y(r+y) = m/y
true only when y=r. insufficient

IMO E

Thanks again for all the work. I understand t1 but I couldn't simplify this. How do you add x / y and 40-y / 1.25x to get 160-y / 5x ? 1.25x x x turns into x^2?! Thank you so much.

cans wrote:4th:
y miles: x m/h and 40-y: 1.25x
t1 = y/x + (40-y)/1.25x = 160-y / 5x
t2 = x m/h for 40 miles. means t2 = 40/x hr.
t1 is what percent of t2?? let p%
then t1 = p/100 * t2 -> (160-y)/5x = p/100 * 40/x
p = (160-y)/2

B) gives value of y. thus sufficient
IMO B

I found it came out to 160+y / 5x ? Whatever, I get it. Thanks so much.

myfish wrote:Thanks again for all the work. I understand t1 but I couldn't simplify this. How do you add x / y and 40-y / 1.25x to get 160-y / 5x ? 1.25x x x turns into x^2?! Thank you so much.

cans wrote:4th:
y miles: x m/h and 40-y: 1.25x
t1 = y/x + (40-y)/1.25x = 160-y / 5x
t2 = x m/h for 40 miles. means t2 = 40/x hr.
t1 is what percent of t2?? let p%
then t1 = p/100 * t2 -> (160-y)/5x = p/100 * 40/x
p = (160-y)/2

B) gives value of y. thus sufficient
IMO B

Dear myfish,
Time for a basic formula:
if a/b = c/d = e/f = K then K = (a+c+e)/(b+d+f) or K = (a-c-e)/(b-d-f)

Please verify this with example of 1/2 , 2/4 , 3/6

myfish wrote:Thank you so much for all the amazing help. This one is a bit unclear since GMATPrep software says B is the right answer.

cans wrote:answer to 3rd:
m/r =x/y ????
A) m/y = x/r or mr=xy. Insufficient
B) m+x / r+y = x/y
so we have to check whether (m+x)/(r+y) = m/r
insufficient
x,y are different..

A&B) x=mr/y
thus (m+x)/(r+y) = (m+ (mr/y))/(r+y) = m (r+y) / y(r+y) = m/y
true only when y=r. insufficient

IMO E