I've been following this thread for a while now, thought it's about time I say hello. Great info all together.
I want to share something, while I was doing the test I had the papers sent to me as xps files which I couldn't do anything about. If you need to view them then the only good way you can work them out is by converting them to pdf. I followed a guide called xps to pdf.
Now if you need to edit the files then you would probably need some sort of OCR and then convert it to a document first, then you can answer on the page and do the exercise. Hopefully that will save you some time because it took me a little bit of time to find out.
For Q.1., OA is (A)
For Q.2., OA is (C)
Yes Q2 should be C and not B.
Did it take Pei more than 2 hours to walk a distance of 10 miles along a certain trail (1 mile = 1.6 kilometers rounded to nearest tenth)
1.Pei walked this distance at an average rate of 6.4 kilometers per hour
2. On average, it took Pei more than 9 minutes per kilometer to walk this distance
Did it take Pei more than 2 hours to walk a distance of 10 miles along a certain trail (1 mile = 1.6 kilometers rounded to nearest tenth)
1.Pei walked this distance at an average rate of 6.4 kilometers per hour
2. On average, it took Pei more than 9 minutes per kilometer to walk this distance
10miles = 1.6x10=16KMS
1 says speed=6.4kmph
Distance we know is 16kms
Therefore= Speed = Distance/time
or time = Distance/Speed = 16/6.4 > 2hrs
1 is sufficient.
2. it takes more than 9 minutes per kms.
Lets say 9.1minutes per kms.
Total time to cover 16kms = 9.1 x 16=145.6 minutes,which is greater than
120 mins i.e. 2 hrs.
2 is sufficient,
Hence both 1 and 2 are sufficient.
Whats the OA?
Could somebody help me solve the attached problem ? Sorry for attachment, but since PG does not allow me to draw a sketch, I had not other go.
After reviewing, I can solve it with just Statement I but can't with just Statement II
@ someshvar
as you have already mentioned that from statement I we get d result
from statement II angle(pqr)+angle(prq)=150 so definately angle(qpr)=30 this gives us statement I only from which the problem is solvable so both are sufficient
Could somebody help me solve the attached problem ? Sorry for attachment, but since PG does not allow me to draw a sketch, I had not other go.
After reviewing, I can solve it with just Statement I but can't with just Statement II
The second statement says PQR + PRQ =150. ------- eq (1)
You know from the figure that PRQ + PRS = 180. (linear pair of angles) --eq(2)
substitute value of PRQ from eq(2) in eq(1).
PQR + 180 - PRS = 150
PRS - PQR=30
I guess that solves it!
Anuj
Anuj, I must say you have a sharp brain.:wow: Thanks again.
The second statement says PQR + PRQ =150. ------- eq (1)
You know from the figure that PRQ + PRS = 180. (linear pair of angles) --eq(2)
substitute value of PRQ from eq(2) in eq(1).
PQR + 180 - PRS = 150
PRS - PQR=30
I guess that solves it!
Anuj
Could somebody help me solve the attached problem ? Sorry for attachment, but since PG does not allow me to draw a sketch, I had not other go.
After reviewing, I can solve it with just Statement I but can't with just Statement II
Mate Not able to download this. Dont know why :'(
I wish, I knew it !!! However, I can send it by e-mail , PM me your e-mail address.
aksh4645 SaysMate Not able to download this. Dont know why :'(
Did it take Pei more than 2 hours to walk a distance of 10 miles along a certain trail (1 mile = 1.6 kilometers rounded to nearest tenth)
1.Pei walked this distance at an average rate of 6.4 kilometers per hour
2. On average, it took Pei more than 9 minutes per kilometer to walk this distance
going by the 1st statement the time taken is 16/6.4 hr
by the 2nd statement the time taken is > 12/5 hr
thus by using each statement individually we can say if Pie took more than 2 hours or not .
What fraction of this year's graduating students at a certain are males??
1. Of this year's graduating students, 33 percent of the males and 20 percent of the females transferred from another college
2. Of this year's graduating students, 25 percent traferred from another college
Lets assume total no of students graduating is 100
No of males graduating is x and females will be (100-x)
Statement 1
33% of males i.e 33x/100 and 20% of females i.e 20*(100-x)/100 are from other college. Not sufficient to answer fraction of males graduating
Statement 2
25 % of graduating students are from other college. Not sufficient.
From Statement 1 and 2
Statement 1: No of graduating students transferred from other college = 33x/100 + 20*(100-x)/100
Statement 2: No of graduating students transferred from other college =25
Comparing 1 and 2
33x/100 + 20*(100-x)/100 = 25
13x = 500
x = 500/13
So we can find the fraction of male strudent graduating which is 5/13. So soln is Both statements are together sufficient to answer.
if A bought several pencils and each pencil is either 23 Cents or 21 Cents, how many 23 Cents did A bought?
1. A bought total 6 pencils
2. total value of the pencils A bought was 130 Cents
Guys please help.....
Lets assume there are 2 kinds of pencil X and Y. X of 23 cents and Y of 21 cents.
Statement 1
Its not sufficient
Statement 2
Total cost is 130
so 21X + 23Y = 130
only soln is X=4 and Y=2. Hence statement 2 is sufficient.
Lets assume there are 2 kinds of pencil X and Y. X of 23 cents and Y of 21 cents.
Statement 1
Its not sufficient
Statement 2
Total cost is 130
so 21X + 23Y = 130
only soln is X=4 and Y=2. Hence statement 2 is sufficient.
But how come we guess like this...
is it just a wild guess or there is some way to tackle this type??
Its not a guess but its very much logical, If you see the equation 21x + 23Y=130 you can write is 21(X+Y) + 2Y=130.
Now try this for diff values of X+Y, as sum is 130, 21(X+Y) should be less than 130 and and arnd 130 only as 2Y is a small figure. try for X+Y = 5, then 21(X+Y) will be 105 but 2Y cant be 25.
Now try with X+Y=6, 21(X+Y) will be 126 and 2Y should be 4 so its a logical with sm thinking 😃
Its not a guess but its very much logical, If you see the equation 21x + 23Y=130 you can write is 21(X+Y) + 2Y=130.
Now try this for diff values of X+Y, as sum is 130, 21(X+Y) should be less than 130 and and arnd 130 only as 2Y is a small figure. try for X+Y = 5, then 21(X+Y) will be 105 but 2Y cant be 25.
Now try with X+Y=6, 21(X+Y) will be 126 and 2Y should be 4 so its a logical with sm thinking :)
that is really logical.....Thanks a lot
Hello Friends,
I need an urgent help, My gmat is in 10 days....though i have been performing well in quant but I always get stuck in Coordinate geometry, i have done Coordinate form Manhattan but I still feel that it is not sufficient as per some of the OG questions level.
Please share some material.....
Guys can u please help me in this...I have added the attachment
How come this can not be solved by B??