gmatophobia wrote:
PS Question 1 - Jan 03 Running at their respective constant rates, Machine X takes 2 days longer to produce w widgets than Machine Y. At these rates, if the two machines together produce (5/4)w widgets in 3 days, how many days would it take Machine X alone to produce 2w widgets? (A) 4 (B) 6 (C) 8 (D) 10 (E) 12 Source:
Official Guide | Difficulty: Hard
There are probably better ways to do it, but:
From stem we can derive that Rate(x)=w/t+2 and Rate(y)=w/t
now
(w/t+2)+(w/t))3=5w/4
2t+2/(t^2+2t)=5/12
12(2t+2)=5(t^2+2222t)
0=5t^2-14t-24
0=(5t+6)(5t-4)
t=-6/5 or t=4
Clearly, rates cannot be negative so t=4
Plugging back into first equation, it takes X (4+2)=6 days to produce W widgets, therefore it must take 12 to produce 2w.
For tricky algebra questions like this, I recommend using the answer choices to guide the way you pick numbers
gmatophobia wrote:
DS Question 1 - Jan 03 The median age of 15 children in group A was 12. The median age of 19 children in group B was 9. Two children - the youngest and the oldest - in group B were to group A. What was the new median age of the children in group A? (1) The youngest children in group B was 7 years old. (2) The oldest child in group B was 12 years old. Source:
Experts Global | Difficulty: Hard
Did very little calculations and went by pure logic:
1) Even if we know the age of the youngest child, we cannot conclude how the set will change.
We could have, for example, everyone being the age 12 and the median would be unchanged.
But it could also be the case that element 8 in the set was 12, and element 9 in the set was 15. If that is the case, the new median cannot be determined. Ins.
2) If the oldest of B was 12, then we have added 1 more element in Set A which is the same as the median. So no matter what, we will have the original 12 in position 8 in the first set, and the second 12 will be, in the worst case, position 9. The median will therefore always be 12 after the change in set A.
Sufficient.
B
mysterymanrog wrote:
Did very little calculations and went by pure logic: 1) Even if we know the age of the youngest child, we cannot conclude how the set will change. We could have, for example, everyone being the age 12 and the median would be unchanged. But it could also be the case that element 8 in the set was 12, and element 9 in the set was 15. If that is the case, the new median cannot be determined. Ins. 2) If the oldest of B was 12, then we have added 1 more element in Set A which is the same as the median. So no matter what, we will have the original 12 in position 8 in the first set, and the second 12 will be, in the worst case, position 9. The median will therefore always be 12 after the change in set A. Sufficient. B
Note that this logic is only possible because A is originally a set with an odd number of elements - the element 12 must be in the actual set itself for it to be the median. If it the number of elements were even, you could not use this logic