Around the World in 80 Questions (Day 4): Of the retailers surveyed

Imo A should be the answer as statement 2 doesn't fit any condition we may make out of this problem but statement 1 fits the one where 15% owned just an online store, 50 both physical and online, 25 physical

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solve using 2x2 matrix
ON - online ; NON not online
PS - physical ; NPS not physical
--- ON---NON---Total
PS-- - --- x-------75
NPS-- --- ---- ---25
tota-- 65 --- 35 ---- --- 100
find x

#1

15 percent of the retailers owned an online store but not a physical store.
--- ON---NON---Total
PS-- -50 --- 25-------75
NPS-- 15 --- 10 ---- ---25
tota-- 65 --- 35 ---- --- 100

sufficient

#2
The number of the retailers who owned both an online and a physical store was five times the number of the retailers who did not own either an online or a physical store.

--- ON---NON---Total
PS-- - 5a --- (35-a) -------75
NPS-- --- a---- ---25
tota-- 65 --- 35 ---- --- 100
solve for a we get
4a=40 ; a = 10
sufficient
--- ON---NON---Total
PS-- -50 --- 25-------75
NPS-- 15 --- 10 ---- ---25
tota-- 65 --- 35 ---- --- 100

OPTION D is correct

Of the retailers surveyed, 65 percent owned an online store, and 75 percent owned a physical store. What percent of the retailers owned a physical store but did not own an online store?

(1) 15 percent of the retailers owned an online store but not a physical store.

(2) The number of the retailers who owned both an online and a physical store was five times the number of the retailers who did not own either an online or a physical store.

Online = 65%
Physical = 75%
Case I:
15% owned online but not physical
Thus owning both = 65 – 15 = 50%
And owning only physical = 75 – 50% = 25%


Case II:
Online + physical = 5* none
None = x
Thus 65 + 75 + x – 5x = 100%
Thus x = 50% and percent owning only physical = 25%

Thus D - both are sufficient

My Answer: A, Statement (1) alone is sufficient.

Statement (1): If 15 percent of retailers owned an online store but not physical store based on information given in the question, we can conclude 50% of retailers own both physical and online stores and 25% of retailers own only physical stores. Hence, sufficient.

Statement (2): We can't make predictions of any other percentages based on this information. Hence, insufficient.

Given: Of the retailers surveyed, 65 percent owned an online store, and 75 percent owned a physical store.
Asked: What percent of the retailers owned a physical store but did not own an online store?

	Online Store	Not Online Store	Total
Offline Store			75%
Not Offline Store			25%
Total	65%	35%	100%

(1) 15 percent of the retailers owned an online store but not a physical store.

	Online Store	Not Online Store	Total
Offline Store	65%-15%=50%	75%-50%=25%	75%
Not Offline Store	15%	25%-15%=10%	25%
Total	65%	35%	100%

The percent of the retailers owned a physical store but did not own an online store = 25%
SUFFICIENT

(2) The number of the retailers who owned both an online and a physical store was five times the number of the retailers who did not own either an online or a physical store.

	Online Store	Not Online Store	Total
Offline Store	5x	75%-5x=35%-x	75%
Not Offline Store		x	25%
Total	65%	35%	100%

75%-5x = 35%-x
4x = 40%
x =10%

	Online Store	Not Online Store	Total
Offline Store	50%	25%	75%
Not Offline Store	15%	10%	25%
Total	65%	35%	100%

The percent of the retailers owned a physical store but did not own an online store = 25%
SUFFICIENT

IMO D

Of the retailers surveyed, 65 percent owned an online store, and 75 percent owned a physical store. What percent of the retailers owned a physical store but did not own an online store?

(1) 15 percent of the retailers owned an online store but not a physical store.
retailers that owned both online and offline = 65-15 =50
therefore, percent of the retailers owned a physical store but did not own an online store = 75-50 : 25

sufficient.


(2) The number of the retailers who owned both an online and a physical store was five times the number of the retailers who did not own either an online or a physical store.
let The number of the retailers who owned both an online and a physical store =x
x=5(100 - (75+65-x))
x=5(x-40)
4x=5*40
x=50
therefore, percent of the retailers owned a physical store but did not own an online store = 75-50 =25.

Sufficient

Therefore D

Given: Of the retailers surveyed, 65 percent owned an online store, and 75 percent owned a physical store. What percent of the retailers owned a physical store but did not own an online store?

(1) 15 percent of the retailers owned an online store but not a physical store.

(2) The number of the retailers who owned both an online and a physical store was five times the number of the retailers who did not own either an online or a physical store.

Solution:

Let Total number of retailers = 100
Let Retailers who own online store but not physical store = x
Let Retailers who own physical store but not online store = y
Let Retailers who own both online and physical store = p
Let Retailers who own neither online nor physical store = q

Now per given information:
x + p = 65 ------ (1)
y + p = 75 ------ (2)
Also x + y + p + q = 100 ------- (3)

We need to find value of y


Option 1: 15 percent of the retailers owned an online store but not a physical store.
Here x is given i.e. x = 15
Subtract equation 1 from 2
y + p - x - p = 75 - 65
y - x = 10
y - 15 = 10
hence y = 25 i.e. 25 percent

SUFFICIENT


Option 2: The number of the retailers who owned both an online and a physical store was five times the number of the retailers who did not own either an online or a physical store.
Here relation between p and q is given i.e. p = 5q
From eq. 3
x + y + p + q = 100
x + y + 5q + q = 10
x + y + 6q = 100 ------- (4)
Now adding eq. (1) & (2)
x + p + y + p = 140
x + y + 2p = 140
x + y + 10q = 140 ------ (5)

Subtract eq. (4) from (5)
4q = 40
q =10
Hence p = 50

Substituting value of p in equation 2
y + p =75
y + 50 = 75
hence y = 25 i.e. 25 percent

SUFFICIENT

Correct answer is choice D

The answer is D
Statement 1 gives the individual value of online store i.e 15. From that we can deduce that the number of retailers who own both is 50. Hence you can subtract 75-50 = 20. This is the value of those retailers who only own a physical store.
Hence Sufficient.
Statement 2 says that Both= 5( Neither)
We can assume 100 to be the total number of retailers. so, 100= 65+ 75-5N +N ( 100= online store+ physical store-both + neither)
We get Neither to be = 10. Hence B=5(N) which comes up to the value 50. By finding the both Value you can subtract from Total value of Physical store and get 20. Hence Sufficient

we can use table or overlapping set to solve this question. I will use table. We have to find the yellow cell.

(1) 15 percent of the retailers owned an online store but not a physical store.

As we know that 15% is the overlap between online and the physical store. The percentage of the overlap (yellow region) is 65-15 = 50.

Sufficient.

(2) The number of the retailers who owned both an online and a physical store was five times the number of the retailers who did not own either an online or a physical store.

Let the percentage of neither = x.

Using the table shown we can find that 75 - 5x + x = 35

We will have an unique value. Hence, this information is also sufficient.



IMO D

Of the retailers surveyed, 65 percent owned an online store, and 75 percent owned a physical store. What percent of the retailers owned a physical store but did not own an online store?
online not online Total
physical x 75
not physical 25
Total 65 35 100

(1) 15 percent of the retailers owned an online store but not a physical store.
online not online Total
physical y x 75
not physical 15 25
Total 65 35 100
we know y by 65-15 =50, and we know x = 75-y --> 75-50

(2) The number of the retailers who owned both an online and a physical store was five times the number of the retailers who did not own either an online or a physical store.


online not online Total
physical 5y x 75
not physical y 25
Total 65 35 100
try substitution number y= 5, 5y=25 then x will 50 --> 50>35 wrong
y=10, 5y = 50 then x =25 25<35 ok
y=15, 5y =75 x =0 but 15+0 = 15 not equal to 25 wrong.

Hence the answer is D

using Venn:
a+x=65, b+x=75
a+b+x+y=100
b=?
S2) x=5y, (NS)
S1) a=15 gives x=50 & b=25 (%) (Suff)
Hence A)

(1) 15 percent of the retailers owned an online store but not a physical store.

so 50% =65-15 on both stores. Hence 25% own only physical stores. Hence sufficient. A

(2) The number of the retailers who owned both an online and a physical store was five times the number of the retailers who did not own either an online or a physical store.
x=neither
65+ 75-5x+x=100
from this we can find out only physical stores. Hence sufficient.
so D

Of the retailers surveyed, 65 percent owned an online store, and 75 percent owned a physical store. What percent of the retailers owned a physical store but did not own an online store?

(1) 15 percent of the retailers owned an online store but not a physical store.

(2) The number of the retailers who owned both an online and a physical store was five times the number of the retailers who did not own either an online or a physical store.


Please check attached solution.
Answer is option D. Each statement alone is sufficent.

Please follow the table below for the solution :

OS = online store
PS = physical store
X in- front on OS and PS means does not have.

Of the retailers surveyed, 65 percent owned an online store, and 75 percent owned a physical store. What percent of the retailers owned a physical store but did not own an online store?

(1) 15 percent of the retailers owned an online store but not a physical store.
Start with this 15% of the retailers assumed to be 100 own OS but X PS.
Therefore 15 is available in place "3" of the table this means that 65 - 15 = "1" in the table which is equal to 50.
Now, 75 - 50 = "2" which denotes the percentage of people who had a PS but not a OS.
"2" = 25.
Hence A is sufficient.
Cross B,D,E.

(2) The number of the retailers who owned both an online and a physical store was five times the number of the retailers who did not own either an online or a physical store.
Start with 2)
OS + PS = 5 X OS+X PS
let X OS+X PS = x so x/100 * 100 = X%
So, it means OS + PS = 5x so 5x/100*100 = 5x%
We know X OS "2" value is common.
75 - 5x = 35 - x
x = 10
this means 1 = 50 and therefore 2 = 25

Hence 2 is also sufficient.

Therefore IMO D.

65 percent owned an online store, and
75 percent owned a physical store

(1) 15 percent of the retailers owned an online store but not a physical store.
Only online store = 15 percent
Retailers those have both Online store and physical store = (65-15) = 50 percent
retailers owned a physical store but did not own an online store = (75-50) = 25 percent

(2) The number of retailers who owned both an online and a physical store was five times the number of retailers who did not own either an online or a physical store.
Insuff

Combined
Not necessary

Ans A

This question is from Venn diagram. I wish I could draw here. But let us see how can we explain it without drawing.
Okay, so here it goes. Look at the attached diagram.
Assume A represents online store (65%) and B represents physical store (75%). X represents neither.
1) This tells us about the area of A barring the portion that falls under both A and B. We can calculate the value that lies in both easily and hence we can calculate percent of the retailers that owned a physical store but did not own an online store. Sufficent.
2) This tells us that A and B both area is 5 times x. Hence we have equation: 65+75-5x+x=100. We can calculate x and hence we can calculate percent of the retailers that owned a physical store but did not own an online store. Sufficent.

Answer is D!

From the question, we have 3 equations.
N + O + PO + P = 100%
O + PO = 65%
PO +P = 75%
where N is the retailers who did not own either an online or a physical store, O is the retailers who an online store but did not own a physical store, P is the retailers who own a physical store but did not own online store, PO is the retailers who own both an online store and a physical store.
P = ?
(1) O = 15%. Then we get PO from the second equation and P from the third equation. Sufficient.
(2) PO = 5N. The we replace all three equations with the following equations:
N + O + 5N + P = 100%
O + 5N= 65%
5N+P = 75%
We have 3 unknow and 3 equations. We can solve the equations to get P. Sufficient.
Option D is the answer.