A new tower has just been built at the Verbico military

Could someone please show the complete solution to this problem? I am getting an answer that is not listed in the answer choices. Please help. TIA.

Let logically split this problem.

Old: number of beds = x. x/3 occupied
New: number of beds = 2x (overall 3x). 2x/5 occupied. So, 8x/5 unoccupied.

x/3 now moved from old to new -> unoccupied beds in new will be = 8x/5 - x/3 = 19x/15

So, fraction of unoccupied in new -> # of unoccupied / total in new = (19x/15) / 2x = 19/30 -> D

Ans is D

let x be the no of beds in the original hospital
so 2x is the no of beds in the new hospital

now x/3 and 2x/5 are occupied. so total is 11x/15

so no of unoccupied beds = 2x-11x/15 = 19x/15

so the fraction is 19x/15/2 = 19x/30

definitely a wordy question.

D as correct answer.
remember that the now the in total it can hold 3times as many people including the old building. Then set your own number of beds for simplicity 15 (old) 30 (new)
1/3 of 15 is 5 and 1/5 of 30 is 6.... 5+6 / 30 is your fraction of beds filled... you must subtract 11 from 30 to get the answer of how many are not filled. 19/30.

Old Hospital = x

New Hospital = 2x (After addition of new hospital, capacity = 3x which is thrice of the old hospital)

Old Hospital Occupancy \(= \frac{x}{3}\)

New Hospital Occupancy\(= \frac{2x}{5}\)

After transfer from old to new hospital,

New Hospital Occupancy\(= \frac{2x}{5} + \frac{x}{3} = \frac{11x}{15}\)

Vacancy in New Hospital \(= 2x - \frac{11x}{15} = \frac{19x}{15}\)

Fraction \(= \frac{19x}{15}* \frac{1}{2} = \frac{19x}{30}\)

Answer = D

Check the working in figure.

Answer: Option D

Forget variables and formulas, try to pick some smart numbers:
Before new tower: 15
After new tower: 15*3=45 (OVERALL ! means new tower + old wing)

New tower: 30 and 20% are occupied = 6
Old wing: 15 (as before renovation) and \(\frac{1}{3}\)are occupied = 5 and then this amount is transfered to the new tower -> 6+5=11 as of now there are 11 beds occupied in the new tower and 30-11=19 beds free, what brings us to our solution\(\frac{19}{30}\) are unoccupied

A new tower has just been built at the Verbico military hospital; the number of beds available for patients at the hospital is now 3 times the number available before the new tower was built. Currently, 1/3 of the hospital's original beds, as well as 1/5 of the beds in the new tower, are occupied. For the purposes of renovating the hospital's original wing, all of the patients in the hospital's original beds must be transferred to beds in the new tower. If patients are neither admitted nor discharged during the transfer, what fraction of the beds in the new tower will be unoccupied once the transfer is complete?

A. 11/30
B. 29/60
C. 17/30
D. 19/30
E. 11/15

let x=original beds
2x=new tower beds
(x/3)+(2x/5)=11x/15 occupied beds in new tower after transfer
2x-(11x/15)=19/15*x unoccupied beds in new tower after transfer
(19/15*x)/2x=19/30
D

Hi All,

I'm a big fan of TESTing Values (BrainLab's explanation), but this question can also be solved with algebra….

We're told that adding a new tower to an existing hospital allows fore 3 times the number of available beds that were there before the tower was built:

Just the hospital = X beds
Hospital + new tower = 3X beds

Just the tower = 3X - X = 2X beds

We're told that 1/3 of the original hospital beds and 1/5 of the tower beds are occupied:

1/3 of X beds + 1/5 of 2X beds = X/3 + 2X/5 = 5X/15 + 6X/15 = 11X/15 occupied beds

Next, we're told that all of the patients in the original hospital must be transferred to the tower. Since the tower has 2X beds in it, the tower is now…..

[11X/15]/2X full = 11/30 full

The question asks for the fraction of the beds in the tower that are UNOCCUPIED:

30/30 - 11/30 = 19/30 UNOCCUPIED

Final Answer:

GMAT assassins aren't born, they're mad,
Rich

Let total number of beds in the hospital before the new tower =\(x\)
Now , the total number of beds available after the new tower was built = \(3x\)
therefore the total number of beds in the new tower = \(3x - x = 2x\)

now total number of occupied beds from the original beds(Before the new tower was built)
= \(\frac{1}{3}x\)

now total number of occupied beds from the new tower
= \(\frac{1}{5}2x\)

number of beds unoccupied from the new tower
= \(2x\)- \(\frac{2}{5}x\) = \(\frac{8}{5}x\)


now, numbers of beds left unoccupied in new tower after we shift the patients from old building to new tower
= \(\frac{8}{5}x\) -\(\frac{1}{3}x\)
= \(\frac{19}{15}x\)

Now, we have total number of unoccupied beds in the new tower(\(\frac{19}{15}x\)), as well as the total number of beds(occupied and unoccupied) in the new tower(\(2x)\)


therefore the answer is = \(\frac{19x}{15 *2x}\) = \(\frac{19}{30}\)
Option D

We can assume there are 30 beds available in the original hospital wing, and thus there are 60 beds available in the new tower (notice that the total number of beds in the two locations is 90, which is 3 times the number of beds in the original hospital wing). We know that 1/3 x 30 = 10 beds in the original wing and 1/5 x 60 = 12 beds in the new tower are occupied, and the 10 patients from the original wingr are transferred to the new tower. So the new tower now has 12 + 10 = 22 patients, and thus the number of unoccupied beds is 60 - 22 = 38. Therefore, the fraction of the beds in the new tower that are unoccupied after the transfer is 38/60 = 19/30.

Answer: D .

Given:
1. A new tower has just been built at the Verbico military hospital; the number of beds available for patients at the hospital is now 3 times the number available before the new tower was built.
2. Currently, 1/3 of the hospital's original beds, as well as 1/5 of the beds in the new tower, are occupied.
3. For the purposes of renovating the hospital's original wing, all of the patients in the hospital's original beds must be transferred to beds in the new tower.

Asked: If patients are neither admitted nor discharged during the transfer, what fraction of the beds in the new tower will be unoccupied once the transfer is complete?

Let the number of beds originally be x
Number of beds after a new tower is built = 3x
Number of beds in new tower = 3x-x = 2x

Occupied beds: -
x/3 original beds
2x/5 beds in new tower
Since all patients in original beds are transferred to new tower
Occupied beds in new tower will become = x/3 + 2x/5 = 11x/15
Fraction of beds occupied in new tower = 11x/15 / 2x = 11/30
Fraction of beds unoccupied in new tower = 1 - 11/30 = 19/30

IMO D

A new tower has been built and now the hospital has 3 times the number of beds available. From this piece of information we can conclude that the new wing has twice the number of beds as the old wing. Combined, there are 3x the number of beds.

We're told 1/3 of the old hospital beds are occupied; 1/5 of the new beds are occupied.

What are the number of unoccupied beds if all the beds were moved to the new tower?

Since we know the new tower is twice as big as the old tower, lets simply divide 1/5 by 2 = 1/10

Now, lets combine 1/3 + 1/10 = 11/30 occupied beds.

Therefore 19/30 are unoccupied. Answer is D.

OE:


Since the question does not give any concrete values, use Smart Numbers to create a value for the number of beds.

In problems involving Fractions, it is best to pick numbers that are multiples of the denominators. Therefore, in the problem at hand, pick numbers that are multiples of 3 and 5. Pay attention! The new tower is not itself three times the size of the old wing; the problem states that the capacity of the entire hospital is three times its original value, so the new tower has twice as many beds as the old wing.

Therefore, let there be 15 beds in the hospital's original wing and 30 in the new tower. In this case, 5 of the original beds and 6 of the beds in the new tower are occupied. If the 5 patients in the original beds are transferred to the new tower, 11 of the beds in the new tower will be occupied; therefore, 19 of the new tower's 30 beds will be unoccupied.

Alternatively, this problem can be solved using Algebra, where x represents the number of beds in the new tower. (This is the most judicious choice of variable, because the problem ultimately asks for a fraction of the beds in the new tower.)

(1/5)x + (1/3)(1/2)x =
(1/5)x + (1/6)x =
(6/30)x + (5/30)x =
(11/30)x

Therefore, 11/30 of the beds in the new tower are occupied, so 1 – 11/30 = 19/30 of them are unoccupied.

The correct answer is (D).

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