PLS REVERT

My approach:
Premises:
ALL tulip > ALL maple
SOME sycamore > ALL maple
ALL maple > ALL dogwoods

(A) Some dogwoods = youngest tulip. Wrong ALL tulip > ALL maple > dogwoods
(B) Some dogwoods = youngest sycamores. Not sure, so can't be MUST BE TRUE. For instant. All dogwoods are 1 year old. All maples are 2 years old. All sycamores are older than 1.5 years old but majority are older than 2 years old.
(C) Some sycamores < Oldest dogwoods. Same as (B). Not sure, so can't be MUST BE TRUE
(D) Some tulip < Oldest sycamore. Same as (D).Not sure, so can't be MUST BE TRUE
(E) Some sycamore < youngest tulip. As some sycamore < Oldest of maple (majority of sycamore > oldest maple) -> some sycamore < Oldest of maple < youngest tulip ( as ALL tulip > ALL maple)

So (E) is the answer. How do you think ?

pappueshwar wrote:In the Centerville Botanical Gardens, all tulip trees are older than any maples. A majority, but not all, of the garden's sycamores are older than any of its maples. All the garden's maples are older than any of its dogwoods.

If the statements above are true, which one of the following must also be true of trees in the Centerville Botanical Gardens?
A. Some dogwoods are as old as the youngest tulip trees.
B. Some dogwoods are as old as the youngest sycamores.
C. Some sycamores are not as older as the oldest dogwoods.
D. Some tulip trees are not as old as the oldest sycamores.
E. Some sycamores are not as old as the youngest tulip trees.

OA IS E. CLEAN BOWLED how to solve this

All tulip trees are older than any maples:
(all tulips) > (all maples) ---> (all maples) < (all tulips).

Not all of the garden's sycamores are older than any of its maples:
(not all sycamores) > (all maples) ---> (some sycamores) ≤ (all maples).

Linking together (some sycamores) ≤ (all maples) and (all maples) < (all tulips), we get:
(some sycamores) < (all tulips).

Answer choice E: Some sycamores are not as old as the youngest tulip trees.

The correct answer is E.

awesome awesome awesome expln mitch !!!!

thanks a ton

This is a LSAT Prep Test question:

Mitch has explained it clearly and beautifully. I typed this explanation in a mail and thought posting it here will be of some use.


1. All tulips are older than any maples -- so youngest tulip is older than oldest maple --- 1
2. Majority of sycamores are older than any of maples -- 51 sycamores are older than oldest maples ( also no upper limit defined ) but 49 sycamores are younger than oldest maples ( also no lower limit defined ) --- 2
3. All maples older than dogwoods - Youngest maple is older than oldest dogwood. --- 3

Option A - Some dogwoods are as old as youngest tulip -- WRONG from 3 and 1 none of dogwood is older than any of maple and none of maple is older than any of tulip. Hence clearly no dogwood is older than any tulip.

Option B - WRONG because from 2 and 3 this MAY BE TRUE but not MUST BE TRUE.

Option C - WRONG because from 2 and 3 this MAY BE TRUE but not MUST BE TRUE

Option D - Again WRONG because this MAY BE TRUE but not MUST BE TRUE

Option E - Some sycamores are not as old as youngest tulip trees. Awesome. When it was stated that 51 sycamores are older than maple 49 were definitely younger than maple. What is younger than maple will also be younger than tulips. Hence always correct.

Please advise if 'what is younger than maple will also be younger than tulips' is correct or 'what is younger to maple will be younger to tulips' is ?