A store's monthly sales total declined j percent from October to Novem

Question

A store's monthly sales total declined j percent from October to November. If the store's total sales in December were equal to its total sales in October, by what percent, in terms of j, did the store's sales increase from November to December?

A. j

B. \(\frac{100j}{100+j}\)

C. \(\frac{10,000j}{100-j}\)

D. \(\frac{100j}{100-j}\)

E. \(\frac{j}{1-j}\)

A. j

B.  \frac{100j}{100+j}

C.  \frac{10,000j}{100-j}

D.  \frac{100j}{100-j}

E.  \frac{j}{1-j}

Lampard42 · Answer

First method (plug-in)

S=100 fall=20% so increase is 25% (20*100/80). Only Option E gives 25%

Second method

N=S*(1-J) D=S; Increase is S-S*(1-J)=S-S+SJ=SJ; % Increase = SJ/S*(1-J); S is eliminated to get J/(1-J) IMO E

Archit3110 · Answer

let october sale 10 and nov = 8 so drop by 20%=j
and Decemer = 10 so increase by 25%
find answer which gives j=.2 as 25%
IMO E  \frac{j}{1-j}

CareerGeek · Answer

Let the sales in October = 100

Sales decreased by j % in November 
—> Sales in November = 100 - j

Sales in December should be 100 (as of October)
—> % increase = (Final - Initial)/Initial*100
= (100 - (100 - j))/(100 - j)*100
= j/(100 - j)*100
= 100j/(100 - j)

IMO Option D

Pls Hit kudos if you like the solution

Posted from my mobile device

ScottTargetTestPrep · Answer

We can let sales in October = C, so sales in November are C(100 - j)/100.

We can let n = the percent increase and create the equation:

C(100 - j)/100 x (100 + n)/100 = C

(100 - j)/100 x (100 + n)/100 = 1

(100 - j)(100 + n) = 10,000

10,000 - 100j + 100n - jn = 10,000

100n - jn = 100j

n(100 - j) = 100j

n = 100j/(100 - j)

Answer: D

CrackverbalGMAT · Answer

A store's monthly sales total declined j percent from October to November. If the store's total sales in December were equal to its total sales in October, by what percent, in terms of j, did the store's sales increase from November to December?

A. j

B. \(\frac{100j}{100+j}\)

C. \(\frac{10,000j}{100-j}\)

D. \(\frac{100j}{100-j}\)

E. \(\frac{j}{1-j}\)

The best way to approach this question would be by using plugin values.

Assume that Oct sales be 100 and the % decrease from Oct to Nov is 20 % i.e j =20.

Nov sales = 80% of 100 = 80.

It's given that December sales is same as the total sales in October.

December sales = 100.

We need to find by what percentage store sales increase from Nov to Dec. i.e from 80 to 100.

Percentage increase = (20/80) * 100 = 25 % 
That means if we substitute j as 20 in answer options, we should get 25.

Only option D satisfies this condition.
100j/(100−j) = 100*20/(100-20) = 100*20/80 = 25

Option D is the answer.

Thanks,
  Clifin J Francis,
GMAT SME

luisdicampo · Answer

Deconstructing the Question

Sales decline by  j%  from October to November. 
December sales equal October sales. 
We must find the percent increase from November to December.

Step-by-step

Let October sales be  S .

November sales:

S\left(1-\frac{j}{100}\right)

Let the increase from November to December be  k% .

S\left(1-\frac{j}{100}\right)\left(1+\frac{k}{100}\right)=S

Divide by S:

\left(1-\frac{j}{100}\right)\left(1+\frac{k}{100}\right)=1

Solve:

1+\frac{k}{100}=\frac{1}{1-\frac{j}{100}}

1+\frac{k}{100}=\frac{100}{100-j}

\frac{k}{100}=\frac{j}{100-j}

k=\frac{100j}{100-j}

Answer: 100j / (100 − j)

A store's monthly sales total declined j percent from October to Novem

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A store's monthly sales total declined j percent from October to Novem

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