190 students go to a school bake sale. 95 buy a chocolate chip cookie,

Question

190 students go to a school bake sale. 95 buy a chocolate chip cookie, 75 buy a peanut butter cookie, and at least 12 buy both. What is the least number of students who could have bought neither type of cookie?

A. 10
B. 24
C. 30
D. 32
E. 45

A. 10
B. 24
C. 30
D. 32
E. 45

EgmatQuantExpert · Accepted Answer

Solution

Given:  
 • 190 students go to a school bake sale
• 95 buy a chocolate chip cookie
• 75 buy a peanut butter cookie
• At least 12 buys both cookie

To find:  
 • The least number of students who could have bought neither type of cookie

Approach and Working:   
All the students can be categorised into one of the 4 categories:
 • Students buying only chocolate chip cookie (95 – 12)
• Students buying only peanut butter cookie (75 – 12)
• Students buying both types of cookie (at least 12, assuming exactly 12 as of now)
• Students buying neither types of cookie (n)

Therefore, we can write,
 • 95 – 12 + 75 – 12 + 12 + n = 190
Or, n + 158 = 190
Or, n = 32

We are trying to minimize the value of n, keeping in mind that at least 12 students buy both types
 • If the number is increased from 12, the value of n will also increase
• Therefore, 32 is the least possible value for n

Hence, the correct answer is option D.

Answer: D

Solve the Question of the Week:      eGMAT Question of the Week

alldaytestprep · Answer

190 students go to a school bake sale. 95 buy a chocolate chip cookie, 75 buy a peanut butter cookie, and at least 12 buy both. What is the least number of students who could have bought neither type of cookie?

A. 10
B. 24
C. 30
D. 32
E. 45

A. 10
B. 24
C. 30
D. 32
E. 45

See attached chart for setup. To find the minimum the no CC/no PB value can be we need to MAXIMIZE the values of x and y. That means we need to minimize the value of PB/CC which is 12. Therefore the max x & y can be are 83 and 63 respectively. This gives a minimum value for No CC/No PB of 32.

PKN · Answer

Explanation:-

Please refer the venn diagram affixed herewith.

n=190-(a+b+c), n is minimum when (a+b+c) is maximum.

We have range of c :  12\leq{c}\leq{75}

when c=12, we have a=95-12=83 and b=75-12=63
 so, a+b+c=83+63+12=158

when c=75, we have a=95-75=20 and b=75-75=0
 so, a+b+c=20+0+75=95

Therefore, maximum value of a+b+c is158.

hence, minimum(n)=190-158=32.

The least number of students who could have bought neither type of cookie is 32.

Option: D

ScottTargetTestPrep · Answer

We can use the equation:

#total = #chocolate chip + #peanut butter - #both + #neither

#neither = #total - #chocolate chip - #peanut butter + #both

As we can see from the equation, keeping everything else constant, the number of students who purchase neither kind decreases as the number of students who purchase both kind decreases. Therefore, to minimize #neither, we should minimize #both:

190 = 95 + 75 - 12 + neither

190 = 158 + neither

32 = neither

Answer: D

EMPOWERgmatRichC · Answer

Hi All,

We're told that 190 students go to a school bake sale. Of those students, 95 buy a chocolate chip cookie, 75 buy a peanut butter cookie, and AT LEAST 12 buy both. We're asked for the LEAST number of students who could have bought NEITHER type of cookie. This question is a variation on a standard Overlapping Sets question (although there is a 'twist'; AT LEAST 12 students bought both types of cookie), but we can still use the Overlapping Sets Formula:

Total = (1st group) + (2nd group) - (Both) + (Neither)

Based on the given information, the equation would look like this:

190 = (95) + (75) - (AT LEAST 12) + (Neither) 
190 = 170 - (AT LEAST 12) + (Neither) 
20 = (Neither) - (AT LEAST 12)

To minimize the "neither group", we have to make the "both" group as SMALL as possible. In this case, that would be exactly 12 people...

20 + 12 = Neither 
32 = Neither

Final Answer:  D

GMAT assassins aren't born, they're made, 
Rich

fskilnik · Answer

https://i.ibb.co/N23M1ph/06-Dec18-2x.gif

? = {\left( R \right)_{\min }}

R = 190 - \left( {95 + 75 - {\rm{both}}} \right) = 20 + {\rm{both}}

{\rm{both}} \ge 12\,\,\,\left( {{\rm{given}}} \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,? = 32

This solution follows the notations and rationale taught in the GMATH method.

Regards, 
Fabio.

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

190 students go to a school bake sale. 95 buy a chocolate chip cookie,

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

190 students go to a school bake sale. 95 buy a chocolate chip cookie,

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards