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23 Jul 2019, 08:00
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A certain building has only 2-room and 3-room apartments to rent. If the range of the rental fees of all the apartments in the building is $1,000 and the range of the rental fees for 3-room apartments is $600, what is the range of the rental fees for 2-room apartments?
(1) The lowest rental fee for a 3-room apartment in the building is $200 more than the highest rental fee for a 2-room apartment in the building.
(2) The lowest rental fee for a 2-room apartment is $800 in the building.
(A) Statement 1 alone is sufficient, but statement 2 alone is not.
(B) Statement 2 alone is sufficient, but statement 1 alone is not.
(C) Both statements together are sufficient, but neither alone is sufficient.
(D) Each statement alone is sufficient.
(E) Neither statement is sufficient, even when combined.
(1) The lowest rental fee for a 3-room apartment in the building is $200 more than the highest rental fee for a 2-room apartment in the building.
(2) The lowest rental fee for a 2-room apartment is $800 in the building.
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(A) Statement 1 alone is sufficient, but statement 2 alone is not.
(B) Statement 2 alone is sufficient, but statement 1 alone is not.
(C) Both statements together are sufficient, but neither alone is sufficient.
(D) Each statement alone is sufficient.
(E) Neither statement is sufficient, even when combined.
23 Jul 2019, 08:31
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A certain building has only 2-room and 3-room apartments to rent. If the range of the rental fees of all the apartments in the building is $1,000 and the range of the rental fees for 3-room apartments is $600, what is the range of the rental fees for 2-room apartments?
This is an easy question.
Let's analyze the stem first.
Lets LABEL
2 room Lowest X
2 room Highest Y
3 room Lowest Z
3 room Highest T
Its is given to use that the range combined is 1000
and T-Z = 600 ....1
We are looking for Y - X
Let's analyze the statements now.
(1) The lowest rental fee for a 3-room apartment in the building is $200 more than the highest rental fee for a 2-room apartment in the building.
This gives us Z = Y +200 ....2
This tells us that the highest of the 2 room is still lower than the lowest of the 3 room. Which helps us establish that T - X = 1000 ....3
Y - X = (Puting 2 & 3 here ) Z-200 - T+ 1000 = Z-T +800 . Now put 1 here =T- 600- T + 800 = 200 is the answer.
Hence this statement is correct.
(2) The lowest rental fee for a 2-room apartment is $800 in the building.
This statement is insufficient.
Hence the answer is A.
This is an easy question.
Let's analyze the stem first.
Lets LABEL
2 room Lowest X
2 room Highest Y
3 room Lowest Z
3 room Highest T
Its is given to use that the range combined is 1000
and T-Z = 600 ....1
We are looking for Y - X
Let's analyze the statements now.
(1) The lowest rental fee for a 3-room apartment in the building is $200 more than the highest rental fee for a 2-room apartment in the building.
This gives us Z = Y +200 ....2
This tells us that the highest of the 2 room is still lower than the lowest of the 3 room. Which helps us establish that T - X = 1000 ....3
Y - X = (Puting 2 & 3 here ) Z-200 - T+ 1000 = Z-T +800 . Now put 1 here =T- 600- T + 800 = 200 is the answer.
Hence this statement is correct.
(2) The lowest rental fee for a 2-room apartment is $800 in the building.
This statement is insufficient.
Hence the answer is A.
From the statement we have the following:
Range is Max - Min = 1000
Range for 3 - room = Max3 - Min3 =600
From (1) The lowest rental fee for a 3-room apartment in the building is $200 more than the highest rental fee for a 2-room apartment in the building, we have that:
Min3 - Max2 = 200, then the range will have to be Max3 - Min2 =1000
but since we have Max3 - Min3 =600, Min3-Max2 = 200, the difference between Max2 - Min 2 must be 200 in order to complete the range.
So, sufficient.
From (2) The lowest rental fee for a 2-room apartment is $800 in the building, we have that:
We have the lowest rental fee for a 2 room but we do not anything about the max value of 2, so clearly insufficient.
(A) is our answer
Range is Max - Min = 1000
Range for 3 - room = Max3 - Min3 =600
From (1) The lowest rental fee for a 3-room apartment in the building is $200 more than the highest rental fee for a 2-room apartment in the building, we have that:
Min3 - Max2 = 200, then the range will have to be Max3 - Min2 =1000
but since we have Max3 - Min3 =600, Min3-Max2 = 200, the difference between Max2 - Min 2 must be 200 in order to complete the range.
So, sufficient.
From (2) The lowest rental fee for a 2-room apartment is $800 in the building, we have that:
We have the lowest rental fee for a 2 room but we do not anything about the max value of 2, so clearly insufficient.
(A) is our answer
