In a class of 100 students, more students were born in October than in

Question

In a class of 100 students, more students were born in October than in any other month. What is the least possible number of students who were born in October?

A. 8
B. 9
C. 10
D. 11
E. 12

gmatophobia · Accepted Answer

To find the minimum number of students who were born in October we have to distribute in such a way to ensure that a maximum number of students are born in the remaining months.

Let's assume that 8 students are born in each of the 12 months

Number of students = 8 * 12 = 96

Number of students remaining = 100 - 96 = 4

As October has the highest number of students born, two of the remaining 4 students can be added to October and the remaining two students can be distributed one each to other months.

Hence, minimum number of students born in October = 8 + 2 = 10

Option C

Bunuel · Answer

100/12 = 8.something. So, the least number must be more than 8.

If 9 students were born in October, and 8 students in each of the remaining eleven months, the total would be 9 + 8*11 = 97, which is less than 100. So, the least number must be more than 9.

However, with 10 students born in October, the remaining 90 students can be distributed in the eleven months without any month having 10 or more students. 8 students in eleven months would total 88, so we could make nine months to have 8 students and two months to have 9 students. Therefore, 10 is the least possible number of students who were born in October.

Answer: C.

samdoesthegmat · Answer

I don't understand why 10 is the correct answer here. There is the possibility that the remaining two students could be distributed to one other month (e.g., September), which would make October & September equal, and thus the condition of October having more students born than in any other month would not be true. 8*10 + 10*2 = 100 so this is entirely possible. I chose 11 because that is the only number the certifies none of the other months would match October's number of students.

Unless the question is looking for a response that does not need 100% certainty for the condition to occur? How am I to assume that is the case in future GMAT questions??

Unless the question is looking for a response that does not need 100% certainty for the condition to occur? How am I to assume that is the case in future GMAT questions?

Bunuel · Answer

The question asks to find the least POSSIBLE number of students who were born in October. This number cannot be 9, as explained  here . However, it COULD be 10, if nine months have 8 students and two months have 9 students. Therefore, 10 is the least POSSIBLE number of students who were born in October.

VashYerma · Answer

A makeshift way of doing this could be to check for the multiple of 11, when we multiply 11 by 9, we get 99, which shows us that if october has 10 births, the 11 left could all have 9 (since there is no rule which states it cannot) and therefore we can adjust a few numbers to make a total of 100, however when we only deduct 8 & 9 from 100, we are left with 7 & 8 which give 77 and 88, which do not add up to a 100 in the best case scenario.

DanTheGMATMan · Answer

Good optimization problem. Max the other months and the distribute the remainder to minimize october:

https://www.youtube.com/watch?v=938mwZKKokQ

KarishmaB · Answer

October has most students. To minimize the value allocated to October, distribute as equally as you can (base case) and then give October just a smidge more. 12*8 = 96 so equal distribution will involve all months getting 8 students each and then 4 months getting 1 more student. 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9 October could have 9 but then it doesn't have more than every other month. So October must have 10 students. 8, 8, 8, 8, 8, 8, 8, 8, 8, 10, 9, 9 - This works. Answer (C) Here is another such question: https://youtu.be/_-gLSW0Z4oI

Vidhi12 · Answer

Remember  more students were born in October than in any other month  
 
lets solve using the options
1) 8
12 months-->8 students each
total= 96--> remaining= 4 students
case 1:
all 4 can be added to october 
case 2: 
if we add one student to october and remaining 3 to three different months, October along with 3 other months will have 9 students. thus  more students were born in October than in any other month  condition will not be met (4 months have 9 students, hence October doesnt have the greatest value)

2) 9
we have actually checked for 8 and 9 both in the option above. using the same logic this option can be eliminated

3) 10
If October has 10 students--> remaining=90
90/11 months--->all 11 months will have either 8 or 9 students and since October has 10 it is the highest.

Since we have to find the least possible number we will not check the rest.

Dhruvek · Answer

or we can solve using averages .
for eg if in oct 8 children were born , then 100-8 = 92 is total no of children born in remaining month.
92/11 = 8. something which is the avg which will definitely has a no 9 in one of the month there it will be greater than oct.

option b 100-9 =91
91/11 = 8. something which will still have 9 in one of the month

option c 100-10 =90
90/11 = 8.....which can have 9 in one of the month but oct has 10 so this can be the answer

luisdicampo · Answer

To find the  least possible number  of students born in October, we need to distribute the students as evenly as possible among all 12 months, while keeping October strictly greater than any other month.

Let  x  be the number of students born in October.
There are 11 other months in the year.

The Logic of "Minimizing the Max" 
To make  x  as small as possible, we must make the other months as large as possible.
However, the constraint is that October must have  more  students than any other month.
Therefore, the maximum number of students in any other month is  x - 1 .

Setting up the Inequality 
The sum of students in October plus the maximum possible students in the other 11 months must be at least the total class size (100).

x + 11(x - 1) \ge 100

Expand and solve:
 x + 11x - 11 \ge 100 
 12x - 11 \ge 100 
 12x \ge 111 
 x \ge \frac{111}{12}

Let's calculate the value:
 111 \div 12 \approx 9.25

Since the number of students must be an integer,  x  must be at least 10.

***

Verification 
Let's see if  9  works (Option B):
If October has 9, the max any other month can have is 8.
Total =  9 + (11 	imes 8) = 9 + 88 = 97 .
This is not enough to cover 100 students. So 9 is impossible.

Let's see if  10  works (Option C):
If October has 10, the other months can have up to 9.
Total capacity =  10 + (11 	imes 9) = 109 .
This is enough. We can distribute the 100 students like this:
* October:  10 
* 2 months with  9  students
* 9 months with  8  students
Total:  10 + 18 + 72 = 100 .
Here, October (10) is strictly greater than any other month (9 or 8).

Answer: C

In a class of 100 students, more students were born in October than in

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

In a class of 100 students, more students were born in October than in

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