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Difficulty:
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20%
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Of the 190 bicycle stores in a certain region, 90 percent repair bicycles and 40 percent rent bicycles. If the integer n denotes the number of bicycle stores in the region that both repair and rent bicycles, which of the following indicates all possible values of n ?
A. 38 ≤ n ≤ 76
B. 57 ≤ n ≤ 76
C. 57 ≤ n ≤ 171
D. 76 ≤ n ≤ 95
E. 76 ≤ n ≤ 171
A. 38 ≤ n ≤ 76
B. 57 ≤ n ≤ 76
C. 57 ≤ n ≤ 171
D. 76 ≤ n ≤ 95
E. 76 ≤ n ≤ 171
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gmatophobia
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Joined: 22 Dec 2016
Last visit: 09 Sep 2026
Posts: 3,174
Location: India
Concentration: Strategy, Leadership
Posts: 3,174
MasteringGMAT
Good one!
We can use a 2*2 matrix to plot the information as shown below -
Attachment:
1.jpg [ 33.95 KiB | Viewed 53871 times ]
Minimum value of n
The value of n is minimum when the cell that represents only repairs is maximum. The maximum value that is possible = 114.
Hence minimum value of n = 171 - 114 = 57
Attachment:
2.jpg [ 40.12 KiB | Viewed 53043 times ]
Maximum value of n
The value of n is maximum when the cell that represents only rent is minimum. The minimum value that the cell = 0.
Hence maximum value of n = 76 - 0 = 76
Attachment:
3.jpg [ 36.72 KiB | Viewed 51906 times ]
Option B
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Power of Tiny Gains
1.01^(365) = 37.78
0.99^(365) = 0.03
Power of Tiny Gains
1.01^(365) = 37.78
0.99^(365) = 0.03
Posts: 1
I solved with Venn diagram:
Total = repair + rent + neither - both
<=> 190 = 190*90% + 190*40% + neither - n
<=> 190 = 171 + 76 + neither - n
Keep solving we got n = 57 + neither
So the minimum number is 57 while highest will depend on number of neither
So we got B and C available
We can eliminate C by knowing neither should be less than 10% of repair (19 stores) then if in extreme case, there is no neither, then both should be 57 + 19.
Or by seeing that if n is equal to 171 then the whole function would break since 171 + 76 + neither - n would not equalize 190 anymore since 190 # 76 + neither
So option B
Total = repair + rent + neither - both
<=> 190 = 190*90% + 190*40% + neither - n
<=> 190 = 171 + 76 + neither - n
Keep solving we got n = 57 + neither
So the minimum number is 57 while highest will depend on number of neither
So we got B and C available
We can eliminate C by knowing neither should be less than 10% of repair (19 stores) then if in extreme case, there is no neither, then both should be 57 + 19.
Or by seeing that if n is equal to 171 then the whole function would break since 171 + 76 + neither - n would not equalize 190 anymore since 190 # 76 + neither
So option B
General Discussion
JeffTargetTestPrep
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Expert reply
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MasteringGMAT
We can organize the data into a cross table:
............. | rent......not rent |
------------------------------------------
repair..... | n..................... | 90%
not repair| ....................... | 10%
------------------------------------------
............. | 40%...........60%| 100%
Finding the maximum possible value of n is easy. Clearly, n cannot be greater than the minimum of the corresponding subtotals:
n ≤ min(90%, 40%) = 40%
To determine the minimum of n, we first need to maximize a cell next to n by using the above technique:
............. | rent......not rent |
------------------------------------------
repair..... | n...............60% | 90%
not repair| ....................... | 10%
------------------------------------------
............. | 40%...........60%| 100%
Now we see that:
n ≥ 90 – 60 = 30%
Therefore, n must be in the following interval:
30% ≤ n ≤ 40%
Since the total number of stores is 190, we have:
57 ≤ n ≤ 76
Answer: B
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Posts: 2
Why are we not doing 90-n + n + 40-n = 190 to find the value of n?
Doing so gives me n = 60, but i don't know how to pick an option after that? any help?
Doing so gives me n = 60, but i don't know how to pick an option after that? any help?
Oppenheimer1945
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X=Repair+Rent
Ans is B
Ans is B
Attachments
image_2023-11-08_214759919.png [ 56.72 KiB | Viewed 46176 times ]
Posts: 17
I did not understand how are we calculating the upper limit (76), if anyone could pls explain in laymen terms, would be appreciable!
Posts: 3
harshshandilya
The equation you made is not accurate because 90 and 40 are percentages, and you are equating them with 190, which is the actual number of stores here.
90% of 190 = 171 stores
40% of 190 = 76 stores
So, the correct equation would be: 171-n + n + 76-n = 190. Solving this, you'll get n=57, which is the minimum limit here.
Subsequently, you can try to think about the maximum limit intutively. In a venn diagram, if you had to maximize 'n' or the overlap area, the rent circle would be completely swallowed by the repair circle, which would mean all 76 stores that offer rent also offer repair. Therefore making 76 the upper limit.
Personally, from a test taking view I feel it's slightly challenging and time consuming to think about minimum and maximum with the above approach as it requires a decent amount of visualization and thinking on your feet. In my opinion, the 4 x 4 matrix method offers a much more structured setup to be able to quickly figure the maximum and minimum limits out in this case.
anikpait17
The number of repair shops is 171, and the number of rental shops is 76
The maximum number of shops that do both rent and repair can be only 76 since there are 91 shops more repair shops than rental shops (171-76), and these can only perform repairs
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MasteringGMAT
Total = repairs + rents - both + neither
repairs = 90% of 190 = 171
rents = 40% of 190 = 76
We can PLUG IN THE ANSWERS, which represent the number of stores that both repair and rent.
Since the number that rent = 76, the number that both repair and rent cannot exceed 76.
Eliminate C, D and E.
Option A implies that the least possible value for both = 38.
If into the blue equation above we plug total = 190, repairs = 171, rents = 76, and both = 38, we get:
190 = 171 + 76 - 38 + neither
190 = 209 + neither
-19 = neither
Not viable.
Eliminate A.
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gmatophobia
I also believe that the correct choice is B but maybe you made a mistake throughout your calculations gmatophobia. In the last column of the tables you should find the 90% and 40% of the 190 bicycle stores and not 19. As a result, these outcomes are 171 and 76 respectively.
Let me know about it
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GMAT Focus 2: 615 Q78 V86 DI78
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I found it faster and simpler to fill out a matrix with the percentages. Maximize and minimize the percentage and then find the percent of 190 with both the max and min percentages.
Posts: 47
gmatophobia
if we are assuming the value of rent and repairs to be maximum (76) then the stores left for repairs only will be (171-76= 95) but there are still 19 stores left that are not categorised in any od them then where will we adjust remaining stores?
Posts: 2
anikpait17
The maximum value of stores that would rent would mean all the stores that repaired also rented. Think of it as the venn diagram above where the one circle is inside the other. Think of it more logically than mathematically. 40% of 190 stores rent bicycles. The maximum of that is 40% of the total (190).
Can any one give an example for all the 3 types (Repair Only - a, Rental Only - b, Rental and Repair - n) composition when n=76?
There are three constrains:
A. All three types total volume is 190: a + b + n =190
B. Total repair is 171: a + n = 171
C. Total Rental is 76: b + n = 76
Solve the equation above, n will be 57
However, I found it imposible to make n = 76
When n = 76,
Based on C, b will be 0
Based on A, a will be 114
However, it contradic with the constrain B, which:
a + n = 114 + 76 = 190, instead of 171 as constrain B.
Can any one found all detail a,b,n composition?
Thanks
There are three constrains:
A. All three types total volume is 190: a + b + n =190
B. Total repair is 171: a + n = 171
C. Total Rental is 76: b + n = 76
Solve the equation above, n will be 57
However, I found it imposible to make n = 76
When n = 76,
Based on C, b will be 0
Based on A, a will be 114
However, it contradic with the constrain B, which:
a + n = 114 + 76 = 190, instead of 171 as constrain B.
Can any one found all detail a,b,n composition?
Thanks
MasteringGMAT
Total bicycle stores = 190
90% repair = 90%(190) = 171 stores.
40% rent = 40% (190) = 76 stores
consider a two set venn diagram which is over lapping. The first circle SET 2 denote repair , over lapping part denotes both repair and rent, and the second circle SET 1 denotes rent.
Set 2 + common= 171
Set 1 + common = 76
summing both we get, 171+76 = 247 = I + 2II .
Roman numerals I denote set2, set 1 and II denote common.
But, the entire box denotes 190. I + II = 190.
Subtracting 247-190 = II = 57
Max value of common is 76. With set 1 =0 , common = 76, and set 2 = 95.
Therefore, the range is 57<= common <=76
OPTION B
Attachment:
GMAT-Club-Forum-2hf3uoxj.png [ 9.74 KiB | Viewed 9927 times ]
Adding KarishmaB to the loop.
I used your youtube video to solve.
Maximize overlap: between 90 and 40 the max overlap is 40 : so 40*190 = 76
for Minimization of overlap start with the smallest (40) and do 100-40 = 60 , then 90-60 = 30: so 30*190 = 57.
I used your youtube video to solve.
Maximize overlap: between 90 and 40 the max overlap is 40 : so 40*190 = 76
for Minimization of overlap start with the smallest (40) and do 100-40 = 60 , then 90-60 = 30: so 30*190 = 57.
MasteringGMAT
Yes, this is correct. Here is that YT video if someone wishes to check it out: https://youtu.be/oLKbIyb1ZrI
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