GMAT Club Around The World!!! If the cost per unit of item A increases

Question

If the cost per unit of item A increases by r percent and the profit per unit of item A decreases by s percent, where s > r, then in terms of r and s, by what percent is the ratio of profit per unit to cost per unit decreased?

A.  \frac{r}{s}

B.  \frac{100 (r-s) }{ 100 + r}

C.  \frac{100 (s+r) }{ 100 + r}

D.  \frac{100 (r+s) }{ 100 + s}

E.  \frac{100 (s-r)}{ 100 + s}

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ananya3 · Answer

IMO C

C_{new} = (1+\frac{r}{100}) C 
 P_{new} = (1-\frac{s}{100}) P

percent decrease
= \frac{ old - new }{ old} * 100

=  (P/C - P_{new}/C_{new}) / P/C * 100

=  ((P/C - ((1-s/100)P / (1+r/100)C)) / P/C  * 100

= (1+r/100 - (1-s)/100) / (1+r/100) * 100

=  \frac{r+s}{100} / \frac{(100 + r)}{100} * 100

=  \frac{100 (r+s) }{ 100 +r}

Aks111 · Answer

Given that cost per unit increases by r%. If original cost was c then new cost is  c(1 + \frac{r}{100}) 
The profit per unit decreases by s%. If original profit was p then new profit is  p(1 - \frac{s}{100}) 
Original ratio of 'profit per unit' to 'cost per unit' = p/c
New ratio =  p(1 - \frac{s}{100}) / c(1 + \frac{r}{100})   =  \frac{p(100-s) }{ c(100+r)}

The decrease in ratio = difference / original 
=  \frac{p}{c} (1- \frac{100-s }{ 100+r})  /  \frac{p}{c} 
=  \frac{100+r - (100-s) }{ 100 +r} 
=  \frac{r+s }{ 100 +r}

The percent decrease =  100 * \frac{r+s }{ 100 +r}

Answer: C

Deepakjhamb · Answer

Solution : c) (r+s) *100 / (100+r)

initial ratio = p/x changed new ratio = p(1-s/100) /x(1+r/100) change or decrease % = initial - final /initial

rocky620 · Answer

Let the original cost be C and profit be P

Initial  profit to cost ratio = P/C

Post Changes 
Cost = C+CR/100 => (100C+CR)/100
Profit = P-PS/100 => (100P-PS)/100

profit to cost ratio =   /   =>  (100P-PS)/(100C+CR)

%change in the ratio =100*  / (P/C), cancel out P/C from numerator and denominator

100 *   = 100*  =  100*

IMO Option C

Green2k1 · Answer

Correct Answer C

If the cost per unit of item A increases by r percent and the profit per unit of item A decreases by s percent, where s > r, then in terms of r and s, by what percent is the ratio of profit per unit to cost per unit decreased?

Initial Cost = x
increased by r. Therefore, new cost= x + x*r/100 = x(1+r/100)

Initial Profit = P
decresed by s. Therefore, new profit= P - P*s/100 = P(1- s/100)

Percent ratio decreased = 100* ((P/x) - (P(1-s/100)/x(1+r/100)))/ (p/x)
 = 100(r+s)/(100+r)

manasi05 · Answer

Lets the initial cost price per unit before increase be c
Lets the initial profit per unit before decrease be p
The original ratio of profit per unit to cost per unit = p/c
Cost price after r percent increase in the price = c(1+r/100)
Profit per unit after s percent decrease in the profit = p(1-s/100)
The changed ratio of profit per unit to cost per unit = p(1-s/100) / c(1+r/100) = p(100-s)/c(100+r)
Formula for percentage decrease =  *100
 /p/c *100
= (1 - 100-s/100+r)*100
= (r+s)*100/100+r

So the answer will be C

TarunKumar1234 · Answer

Given,

Type----------Cost--------------Selling price-----Profit
Initially-------x(say)------------y(say)-----------(y-x)
Finally--------x*(1+r%)--------------------------(y-x)*(1-s%)

Initially, profit/cost = (y-x)/x
Finally, profit/cost = {(y-x)*(1-s%)}/{x*(1+r%)}= {(y-x)/x}*{(100-s)/(100+r)}

% decrease in ratio= (Finally-initially)/Initially *100 = [{(y-x)/x} - {(y-x)/x}*{(100-s)/(100+r)}] /(y-x)/x *100
=(100+r-100+s)/(100+r) *100 = 100*(r+s)/(100+r)

So, I think C.

MakSha · Answer

Let the initial cost per unit be c and profit per unit be p. According to the question,

p' = p(1-s/100)
c' = c(1+r/100)

thus, to calculate the desired value:

(p'/c' - p/c)/(p/c) X 100%

= 100((s+r)/(100+r))

IMO, option C is the answer .

Gmat202023 · Answer

Initial Cost of item : x ------- (1)
Initial Profit : y ------- (2)

New cost of item : (100+r/100)x -------- (3)
New profit : (100-s/100)y -------- (4)

percentage decrease in ratio = ((2)/(1))-((4)/(3))/((2)/(1))
Solving it, you get (r+s)/(100+r)

converting it into percentage : multiply by 100

Therefore, 100(r+s)/(100+r)

CYp113 · Answer

I struggled soo much on this one. 
OA from Princeton Review:
This question has variables in the question stem and in the answer choices, so Plug In. There are four variables in the question stem, so choose values for the cost of Item A, the profit of Item A, r, and s. Let cost per unit of Item A = $20, and let profit per unit = $10. Cost increases by r percent, so let r = 10. Profit decreases by s percent, where s > r, so let s = 20.
Original ratio : profit/cost = 10/20 = 1/2
New ratio: profit/cost = (10*0.8)/(20*1.1) = 8/22 
% decrease = (1/2-8/22) / 1/2 * 100 = 300/11
Plug in r and s for choices. C) 100(r+s)/100+r = 100(10+20)/(100+10) = 3000/110 = 300/11

ThatDudeKnows · Answer

CYp113

I typically like Plugging In when there are variables in the question that are repeated in the answer choices, but yikes, did they end up with some unfavorable numbers!! We are dealing with percents, so I don't understand why they'd go with 20 for two of the inputs...let's try some numbers that make it easier...we can see from the answer choices that we never need to actually plug in the original cost, so how about 100 for that as a starting point?

cost = $100
profit = $1
r = 0%
s = 50%

Original profit/cost = 1/100 = 1%
New profit/cost = 0.5/100 = 0.5%
Difference/Original = 0.5/1 = 50%

A.  \frac{0}{50}  Wrong.

B.  \frac{100 (0-50) }{ 100 + 0}=\frac{-5000}{100}  Wrong.

C.  \frac{100 (50+0) }{ 100 + 0}=\frac{5000}{100}  = 50. Keep it.

D.  \frac{100 (0+50) }{ 100 + 50}=\frac{5000}{150}  Wrong.

E.  \frac{100 (50-0)}{ 100 + 50}=\frac{5000}{150}  Wrong.

Answer choice C.

Getting good at picking numbers is a learned skill. Nailing it can make questions like this a whole lot easier, though! It's worth noting that my guess is that the reason Princeton Review didn't use 100 as one of its values is that on percent questions, we typically don't want to use 100 for one of the variables because the answer choices usually have 100 in some of them and we run the risk of ending up with more than one "Keep it" and having to do the whole thing over again. In this case, we can glance at the answer choices and see that there are only two variables in them. We probably want to avoid 100 for either of those, but that's an invitation to use 100 for (at least one) of the others.

ThatDudeKnowsPluggingIn

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GMAT Club Around The World!!! If the cost per unit of item A increases

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