In a hostel, the number of students decreased by 8% and the

Hi,

Assume there are 100 students, price/student = 10, Consumption = 1unit --> Cost = 100*10*1 = 1000
Given scenario: 92 students, price/student = 12, Consumption = x, But cost = 1000

92*12*x = 1000
x = 1000/1104 = 125/138 = 0.9 (approx)
So consumption has reduced by (1 - 0.9)*100 = 10% (approx)

Answer: E

I cannot see this being a GMAT problem - how is one supposed to calculate 1- [1/(.92*1.2)] without a calculator? Even with using fractions I get 13/138 which is also not easily solved.

Hello

My approach is

Name:
S = number of students
P = Price
Q = Quantity of food consumption.
C = total cost

==> C = S*P*Q

One way to solve this kind of question is plug in numbers.
Assume: before we have 100 students, P = $10 ==> C = 100*10*Q = 1000*Q
After: we have 92 students, P = $12 ==> C' = 92*12*Q' = 1104*Q'

==> 1104*Q' = 1000*Q
==> Q' = 1000/1104*Q = (1 - 104/1104)*Q
==> Q' = Q - 104/1104*Q

The red part is the decrease of food consumption. We can guess 104/1104 ~ 9.4% (The ONLY option < 10%) --> E is answer.

Hope it helps.

Agree.. this calculation goes tough. Any simpler way to avoid that 92 & 12 factor?

Vyshak
Any Shortcuts here ?
took me 8 minutes to solve
i solved twice
actually thrice

regards
Stone Cold

Let there be 100 students , price of each food student be 100$
Now 8% number of students reduced = 92, price of food increased by 20% = 120$
x(x is decimal less than 1) be reduced consumption of food, so price of reduced food = 120x

given 100 * 100 = 92 * 120 * x
x = 10000/(92*120) = 1000/1104 = 10/11 (approx)
so how much each student has to reduce = 1 - 10/11 = 1/11 = 9% approx => closest is (E)

just to clear things up, what you did in the second part was:
C = F(new) * (0.92N) * (1.2P)
FNP = F(new) * (0.92N) * (1.2P)
F(new) = FNP / (0.92N) * (1.2P)
F(new) = F / (0.92) * (1.2)
F(new) = 0.906F

correct?

Rule: If the overall/final value is not changing and if one variable is increased by \(\frac{1}{x}\)% then the other variable should decrease by \(\frac{1}{x+1}\)%
for eg: if AB = 100, now if B is increased by \(33.33\)% then by how much percentage should A decrease such that the final value remains the same.
Now, B is increased by \(33.33\)% --> \(\frac{1}{3}\)% --> \(x=3\)
Then if final value is same, i.e; 100, then A should decrease by \(\frac{1}{x+1}\)% = \(\frac{1}{(3+1)}\)% = \(\frac{1}{4}\)% = \(25\)%

To be more clear - let \(A = 20\) and\( B = 5\) --> \(AB = 20*5\) = \(100\)
Now if \(B = 5\) is increased by \(33.33\)% = \(5*1.3333\) = \(6.6665\), then \(A = 20\) should decrease by \(25\)% = \(20*0.75\) = \(15\),
such that \(AB = 6.6665*15\) = \(99.997\) =\(~100\) (same as before)

We can apply the same logic here
Let say, Total consumption = \(T\)
consumption of food / student = \(c\)
total number of students = \(n\)
price per food = \(p\)

Hence,
\(T = c * n * p\)

Let us take \(n * p = F\)

So, \(T = c * F\)

Now, \(n\) is decreased by \(8\)% and \(p\) is increased by \(20\)%, we can calculate overall % change in \(F\) by formula \(A + B + \frac{AB}{100}\)
overall % change in \(F\) = \(20 - 8 - \frac{20(8)}{100}\) = \(10.4\)%

Total consumption(after change), \(T = c' * 10.4F\), where \(c'\) = change in the consumption

Now, let's apply the above logic we learned,

The total consumption \(T\) is same and one variable \(F\)(here) is increased by \(10.4\)% ~ \(10\)%(approximate) = \(\frac{1}{10}\)% = \(x = 10\)

Hence the other variable \(c'\) (here) should decrease by \(\frac{1}{x+1}\)% = \(\frac{1}{(10+1)}\)% = \(\frac{1}{11}\)% = \(9.09\)% ~ \(9\)%

The only answer option closest to \(9\)% = \(9.4\)% Option E

Kudos if find solution helpful

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