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605-655 Level|   Math Related|         
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GMAT Club Olympics 2025 (Day 7): At a summer camp, a class of students 08 Jul 2025, 09:53
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Let total students be t

• For the math session, the students are divided so that one group has 4 students and all other groups have 5 students each.
t=5k+4

• For the biology session, they are divided so that one group has 4 students and all other groups have 6 students each.
t=6p+4

• For the history session, the same students are divided so that one group has fewer than 7 students and all other groups have 7 students each.
t=7q+x, where x<7

From the first 2 conditions, it seems like t= 30a + 4, as it divides by both 5 and 6 and leaves a remainder 4

Say t=34, we have q=4 and x=6. This works based on the options provided


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At a summer camp, a class of students participates in three different sessions: math, biology, and history.

• For the math session, the students are divided so that one group has 4 students and all other groups have 5 students each.
• For the biology session, they are divided so that one group has 4 students and all other groups have 6 students each.
• For the history session, the same students are divided so that one group has fewer than 7 students and all other groups have 7 students each.

Select for Groups of 7 the number of groups that have 7 students each in the history session, and select for Remaining students the number of students in the remaining history group, that would be jointly consistent with the given information. Make only two selections, one in each column.
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GMAT Club Olympics 2025 (Day 7): At a summer camp, a class of students 08 Jul 2025, 09:58
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For math session => no of students = 4+5k
For biology session => no of students = 4+6m
For history session => no of students = <7 + 7n

Since in all sessions number of students remained same
4+5k = 4+ 6m
This gives whole number when k = 6 and m = 5
Total number of students = 34

for history session we will have 6 + 7*4 = 34

Answer:
Groups of 7 = 4
Remaining students = 6

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At a summer camp, a class of students participates in three different sessions: math, biology, and history.

• For the math session, the students are divided so that one group has 4 students and all other groups have 5 students each.
• For the biology session, they are divided so that one group has 4 students and all other groups have 6 students each.
• For the history session, the same students are divided so that one group has fewer than 7 students and all other groups have 7 students each.

Select for Groups of 7 the number of groups that have 7 students each in the history session, and select for Remaining students the number of students in the remaining history group, that would be jointly consistent with the given information. Make only two selections, one in each column.
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GMAT Club Olympics 2025 (Day 7): At a summer camp, a class of students 08 Jul 2025, 10:11
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There are N students in total, each participating in Math, Biology, and History, with equal numbers in each.

Math: One group has 4, the others have 5. = 5m+4
Biology: One group has 4 members, while the others have 6. =6b+4
History: One group has fewer than 7, and the others have 7. =7h+a, a<7

First, we need to find matching possibilities between Math and Biology.

Math:
4, 9, 14, 19, 24, 29, 34, 39, 44, 49, 54, 59, 64.....
Biology:
4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64......

Matching patterns:
4+30k
4, 34, 64....
But since the scenario doesn't allow 4 (as one group is 4, others are 5 in Math and 6 in Biology), the starting numbers are:

34, 64, 94
Now, let's look at History:
34=4x7+6
64=9x7+1
94=13x7+3

Looking at the options:
Only "Groups of 7 =4, Remaining students" fits.
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GMAT Club Olympics 2025 (Day 7): At a summer camp, a class of students 08 Jul 2025, 10:16
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for Math = 4 for one grp and 5 for other groups = 4 + 5x, where x = No of other groups
for Bio = 4 for one grp and 6 for other groups = 4 + 6y, where y = No of other groups
So 5x = 6y
x:y = 6:5
if x = 6, y = 5 then N = 34 here N = total possible students
if x = 12, y = 10 then N = 64

Now for history = <7 for one group and other have 7 students
lets say 7z, z = no of other groups
34-7z < 7
if z = 2 then it will be 20
if z = 3 then 11
if z = 4 then 6 => 4 and 6 pair available in the column
if z = 5 then 1 now it can't go further

hence groups of 7 = 4 and Remaining students = 6
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GMAT Club Olympics 2025 (Day 7): At a summer camp, a class of students 08 Jul 2025, 10:18
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Maths : 4 | 5 ....
Biology : 4 | 6 ....
History : <7 | 7 ....

For Maths, let groups of 5 be a.
For Biology, let groups of 6 be b.
lets equate the total students :
4 + 5a = 4 + 6b
5a = 6b
Therefore , a =6, b=5.
So total students = 4+5*6 = 34

So on dividing 34 for History according to the given constraints we get :
6 | 7 7 7 7

Thus Groups of 7 = 4 and Remaining students = 6.
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GMAT Club Olympics 2025 (Day 7): At a summer camp, a class of students 08 Jul 2025, 10:24
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Bunuel
 


This question was provided by GMAT Club
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At a summer camp, a class of students participates in three different sessions: math, biology, and history.

• For the math session, the students are divided so that one group has 4 students and all other groups have 5 students each.
• For the biology session, they are divided so that one group has 4 students and all other groups have 6 students each.
• For the history session, the same students are divided so that one group has fewer than 7 students and all other groups have 7 students each.

Select for Groups of 7 the number of groups that have 7 students each in the history session, and select for Remaining students the number of students in the remaining history group, that would be jointly consistent with the given information. Make only two selections, one in each column.
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Given the number of students is :
Maths = 5*n + 4 where n is any natural number
Biology = 6*n + 4 where n is any natural number

5 and 6 can have same remainder when the number without the remainder is same that is the number is LCM of 5 and 6 = 30
So the number of student is 30 + 4 = 34.

Therefore, the groups of 7 students would be 4 (ANSWER for 1st column) as 7*4 =28 <34
and the remaining students = 34 - 28 = 6 Answer for 2nd column
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GMAT Club Olympics 2025 (Day 7): At a summer camp, a class of students 08 Jul 2025, 10:38
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Let N be the total number of students in the class.

From the math session description:
The students are divided into groups of 5, with one group having 4 students.
This can be expressed as N=5k+4 for some non-negative integer k.
This means N≡4(mod5).
Possible values for N: 4, 9, 14, 19, 24, 29, 34, 39, 44, 49, 54,...

From the biology session description:
The students are divided into groups of 6, with one group having 4 students.
This can be expressed as N=6j+4 for some non-negative integer j.
This means N≡4(mod6).
Possible values for N: 4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64,...

We need to find a value of N that satisfies both conditions.
N≡4(mod5)
N≡4(mod6)

This implies that N−4 must be a multiple of both 5 and 6. Therefore, N−4 must be a multiple of the least common multiple (LCM) of 5 and 6.
LCM(5, 6) = 30.

So, N−4=30m for some non-negative integer m.
N=30m+4.

Let's find the possible values for N:

If m=0, N=30(0)+4=4.

If m=1, N=30(1)+4=34.

If m=2, N=30(2)+4=64.

Now, let's consider the history session description:
The students are divided into groups of 7, with one group having fewer than 7 students.
This means N=7p+r, where p is the number of groups with 7 students each, and r is the number of students in the remaining group, such that 0<r<7. The condition "one group has fewer than 7 students" implies a remainder r that is not zero, and less than 7.

Let's test the possible values of N we found:

Case 1: N=4

Math: 1 group of 4. (Consistent: 4=5(0)+4)

Biology: 1 group of 4. (Consistent: 4=6(0)+4)

History: 4=7(0)+4.
Here, p=0 (Groups of 7) and r=4 (Remaining students).
This is consistent with "one group has fewer than 7 students" (4<7).

Case 2: N=34

Math: 34=5(6)+4. (Consistent: 6 groups of 5, 1 group of 4)

Biology: 34=6(5)+4. (Consistent: 5 groups of 6, 1 group of 4)

History: 34=7(4)+6.
Here, p=4 (Groups of 7) and r=6 (Remaining students).
This is consistent with "one group has fewer than 7 students" (6<7).

Case 3: N=64

Math: 64=5(12)+4. (Consistent: 12 groups of 5, 1 group of 4)

Biology: 64=6(10)+4. (Consistent: 10 groups of 6, 1 group of 4)

History: 64=7(9)+1.
Here, p=9 (Groups of 7) and r=1 (Remaining students).
This is consistent with "one group has fewer than 7 students" (1<7).

The problem asks to select "the number of groups that have 7 students each in the history session" and "the number of students in the remaining history group, that would be jointly consistent with the given information." Since all three cases (N=4, N=34, N=64) provide jointly consistent pairs, the question implies we should provide one such valid pair. In typical multiple-choice scenarios, one of these would be offered as an option. Without explicit options, the smallest non-trivial valid scenario is often preferred, or any valid scenario works.

Let's choose the values from Case 2 (N=34) as a representative valid solution. This case implies there are actual "other groups" of 5, 6, and 7 students, which fits the phrasing well.

For N = 34:

Groups of 7: 4
Remaining students: 6
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GMAT Club Olympics 2025 (Day 7): At a summer camp, a class of students 08 Jul 2025, 10:45
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Firstly, let's establish what we have:

M = 4 + 5a
B = 4 + 6b
H = <7 + 7c

Here, M=Math, B=Biology, H=History, represent the total number of children at the camp
Also, a,b, and c represent the number of groups that are formed with those children. E.g., For the math class, there is one group with 4 kids, and "a" number of groups with 5 kids each


Now, notice that for each of these classes, the number of kids needs to be the same. So, if we equate M and B, we get


4 + 5a = 4 + 6b -----> 5a = 6b

This means that "a" is a multiple of 6, and "b" is a multiple of 5. Let's take a simple case, and assume a = 6, and b = 5

Number of kids = 4 + 5(6) = 34

Now, the number of groups with 7 kids that can be formed amongst these 34 kids is 4. Since, 7*4 = 28

This will leave us with 6 kids who are part of another group in History. Also, this matches our prompt since this number is <7.

So the answer is:

Groups of 7: 4
Remaining students: 6
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GMAT Club Olympics 2025 (Day 7): At a summer camp, a class of students 08 Jul 2025, 10:53
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At a summer camp, a class of students participates in three different sessions: math, biology, and history.

• For the math session, the students are divided so that one group has 4 students and all other groups have 5 students each.
• For the biology session, they are divided so that one group has 4 students and all other groups have 6 students each.
• For the history session, the same students are divided so that one group has fewer than 7 students and all other groups have 7 students each.

Select for Groups of 7 the number of groups that have 7 students each in the history session, and select for Remaining students the number of students in the remaining history group, that would be jointly consistent with the given information. Make only two selections, one in each column.
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Assume that the number of students in the group is N

For the math session, the students are divided so that one group has 4 students and all other groups have 5 students each.

N = 5m + 4

For the biology session, they are divided so that one group has 4 students and all other groups have 6 students each.

N = 6n + 4

As both represent the same number, we can equate

6n + 4 = 5m + 4

6n = 5m

5, 6 are co-primes. Hence, m = 6 and n = 5 for the equation to hold true.

Number of students = N = 5*6 + 4 = 34

For the history session, the same students are divided so that one group has fewer than 7 students and all other groups have 7 students each.

34 = 7x + y

y < 7

Hence, we can have y = 6

x = 4

Groups of 7 = 4
Remaining Students = 6
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GMAT Club Olympics 2025 (Day 7): At a summer camp, a class of students 08 Jul 2025, 11:00
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Let M = Math, B = Biology, and H = History

• For the math session, the students are divided so that one group has 4 students and all other groups have 5 students each.
So, M = p*5 + 4

• For the biology session, they are divided so that one group has 4 students and all other groups have 6 students each.
So, B = q*6 + 4

• For the history session, the same students are divided so that one group has fewer than 7 students and all other groups have 7 students each.
So, H = r*7 + X, 0 < X < 7

LCM ( 5 , 6 ) = 30

Therefore:
Students = 30 + 4 = 34

34 mod 7 = 6
34/7 = 4 with 6 as remainder

Answer:
Groups of 7 = 4
Remaining students = 6
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GMAT Club Olympics 2025 (Day 7): At a summer camp, a class of students 08 Jul 2025, 11:02
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Maths class: 4 in 1 group + 5 in 6 groups = 34 total
Bio class: 4 in 1 group + 6 in 5 groups = 34 total
History class: x in 1 group + 7 in 4 groups = 34, we will get x=6 as x<7

Column 1: 4 groups of 7 students
Column 2: 6 remaining students
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At a summer camp, a class of students participates in three different sessions: math, biology, and history.

• For the math session, the students are divided so that one group has 4 students and all other groups have 5 students each.
• For the biology session, they are divided so that one group has 4 students and all other groups have 6 students each.
• For the history session, the same students are divided so that one group has fewer than 7 students and all other groups have 7 students each.

Select for Groups of 7 the number of groups that have 7 students each in the history session, and select for Remaining students the number of students in the remaining history group, that would be jointly consistent with the given information. Make only two selections, one in each column.
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GMAT Club Olympics 2025 (Day 7): At a summer camp, a class of students 08 Jul 2025, 11:40
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For the math session, the students are divided so that one group has 4 students and all other groups have 5 students each.
let's assume group = x, then Student = 4+5X
For the biology session, they are divided so that one group has 4 students and all other groups have 6 students each.
let's assume group = y, then Student = 4+6y
For the history session, the same students are divided so that one group has fewer than 7 students and all other groups have 7 students each.
let's assume group = Z, then Student = <7+7Z
As students are the same, then
4+5X = 4+6Y

We can have 4, 34 ,..

let try for history session = 34 = 6+7*4 satisfy the eqaution then group of 7 = 4, remaining. = 6
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GMAT Club Olympics 2025 (Day 7): At a summer camp, a class of students 08 Jul 2025, 11:51
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At a summer camp, a class of students participates in three different sessions: math, biology, and history.

• For the math session, the students are divided so that one group has 4 students and all other groups have 5 students each.
• For the biology session, they are divided so that one group has 4 students and all other groups have 6 students each.
• For the history session, the same students are divided so that one group has fewer than 7 students and all other groups have 7 students each.

Select for Groups of 7 the number of groups that have 7 students each in the history session, and select for Remaining students the number of students in the remaining history group, that would be jointly consistent with the given information. Make only two selections, one in each column.
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At a summer camp, students participate in three different sessions : Math, Biology and History.

Math:

Group 1 has 4 students and the remaining groups each had 5 students. This can be expressed as 4+5m.

Biology:

Group 1 has 4 student and the remaining groups each has 6 students. This can be expressed as 4+6b.

History :

Group1 has LESS THAN 7 students and the remaining groups had 7 students each. LT7 + 7*h


Let’s equate Math and Biology, we get 4+ 5m = 4+ 6b

5*m = 6*b

m: b = 6:5

If m = 6, the number of math students = 4+ 5*6 = 34 students.

b = 5 , then the number of Biology students = 4+6*5 = 34 students.

Thus, history = 34 students = 7*4 + 6

So group 1 has 6 students , which is less than 7.

And, the number of groups with 7 students each = 4 groups.
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GMAT Club Olympics 2025 (Day 7): At a summer camp, a class of students 08 Jul 2025, 12:07
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If n is the number of students:

From math, we know: (n-4) is a multiple of 5
From biology, we know (n-4) is a multiple of 6

Since (n-4) is a multiple of both 5 and 6, it will be a multiple of 30.
So (n-4) will be one of these: 30, 60, 90,120,....
and as a result, n will be one of these: 34,64,94,124,...

To calculate the number of groups of 7 and the number of one smaller group, we can start from 34 and check if there are options available.
So, if we write 34 as an equation of multiples of 7 and the remainder, we have: \(34=(7*4)+6 \)
which means we have 4 groups of 7, and one group of 6. Since these options are available, we found the answer.

4 Groups of 7, and 6 students remaining.


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At a summer camp, a class of students participates in three different sessions: math, biology, and history.

• For the math session, the students are divided so that one group has 4 students and all other groups have 5 students each.
• For the biology session, they are divided so that one group has 4 students and all other groups have 6 students each.
• For the history session, the same students are divided so that one group has fewer than 7 students and all other groups have 7 students each.

Select for Groups of 7 the number of groups that have 7 students each in the history session, and select for Remaining students the number of students in the remaining history group, that would be jointly consistent with the given information. Make only two selections, one in each column.
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GMAT Club Olympics 2025 (Day 7): At a summer camp, a class of students 08 Jul 2025, 12:12
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Maths- 4(1) + 5n
Biology- 4(1) + 6m

Total number of students = 6*5 + 4= 34

History-
(I) Group of 7-> 7*4=28
(II) Remaining students > 34 - 28 = 6 (<7)
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At a summer camp, a class of students participates in three different sessions: math, biology, and history.

• For the math session, the students are divided so that one group has 4 students and all other groups have 5 students each.
• For the biology session, they are divided so that one group has 4 students and all other groups have 6 students each.
• For the history session, the same students are divided so that one group has fewer than 7 students and all other groups have 7 students each.

Select for Groups of 7 the number of groups that have 7 students each in the history session, and select for Remaining students the number of students in the remaining history group, that would be jointly consistent with the given information. Make only two selections, one in each column.
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GMAT Club Olympics 2025 (Day 7): At a summer camp, a class of students 08 Jul 2025, 12:25
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Option B and Option C are the correct answers.

Lets understand the question before trying to solve for it.

The question starts by telling that at a summer camp a class of students participated in three different sessions: Math, Biology, and History. Then it further tells us about the group distribution among the students i.e.:

• For the math session, the students are divided so that one group has 4 students and all other groups have 5 students each.
• For the biology session, they are divided so that one group has 4 students and all other groups have 6 students each.
• For the history session, the same students are divided so that one group has fewer than 7 students and all other groups have 7 students each

So from here we can write them down in equation forms, where x, y and z are the number of groups in Math, Biology and History session respectively.
Like for Match Session = 5x + 4
Biology Session = 6y + 4
History session = 7z + a, where a<7.

Now the questions asks us about the "the number of groups that have 7 students each in the history session" and "the number of students in the remaining history group".

Lets try check and insert the option values into the above equations to get the total number of students and then check whether that number satisfy all the equations or not.
First lets assume Group(z) to be '2' and Remaining students(a) to be '2' as well.
History session => 7z + a => 7*2 + 2 => 16 Students
Math Session => 5x + 4 = 16 => 5x = 12 not possible as number of students can not be in decimal. Eliminated

Lets try another pair z = 2 and a = 4
History session => 7z + a => 7*2 + 4 => 18 Students
Math Session => 5x + 4 = 18 => 5x = 14 not possible as number of students can not be in decimal. Eliminated

Let z = 2 and a = 6
History session => 7z + a => 7*2 + 6 => 20 Students
Math Session => 5x + 4 = 20 => 5x = 16 not possible as number of students can not be in decimal. Eliminated

Now we won't be checking z = 2 and z = 8 or 10 because in the question it was clearly mention that the number to Remaining Students is less than '7'.

Let take z = 4 and a = 2
History session => 7z + a => 7*4 + 2 => 30 Students
Math Session => 5x + 4 = 30 => 5x = 26 not possible as number of students can not be in decimal. Eliminated

Lets try z = 4 and a = 4
History session => 7z + a => 7*4 + 4 => 32 Students
Math Session => 5x + 4 = 32 => 5x = 28 not possible as number of students can not be in decimal. Eliminated

Lets try another pair z = 4 and a = 6
History session => 7z + a => 7*4 + 6 => 34 Students
Math Session => 5x + 4 = 34 => 5x = 30 => x = 6 Groups
Biology Session => 6y + 4 = 34 => 6y = 30 => y = 5 Groups

This pair of Option B and Option C satisfies all the equations and conditions mention in the question. So we can conclude here that these are our final answers.

But lets still try to check others to be double sure of our answer.

Let take z = 6 and a = 2
History session => 7z + a => 7*6 + 2 => 44 Students
Math Session => 5x + 4 = 44 => 5x = 40 => x = 8 Groups
Biology Session => 6y + 4 = 44 => 6y = 40 not possible as number of students can not be in decimal. Eliminated

Lets try z = 6 and a = 4
History session => 7z + a => 7*6 + 4 => 46 Students
Math Session => 5x + 4 = 46 => 5x = 42 not possible as number of students can not be in decimal. Eliminated

Lets take z = 6 and a = 6
History session => 7z + a => 7*6 + 6 => 48 Students
Math Session => 5x + 4 = 48 => 5x = 44 not possible as number of students can not be in decimal. Eliminated

If you want you can check rest of the values as well but it will just waist the time which we will get on the exam because each question has only one answer and others might get disqualified by the conditions mention in the question or due to any other reasons. So from here we can conclude that Option B and Option C are the only options. Thus our answer.


Bunuel
 


This question was provided by GMAT Club
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At a summer camp, a class of students participates in three different sessions: math, biology, and history.

• For the math session, the students are divided so that one group has 4 students and all other groups have 5 students each.
• For the biology session, they are divided so that one group has 4 students and all other groups have 6 students each.
• For the history session, the same students are divided so that one group has fewer than 7 students and all other groups have 7 students each.

Select for Groups of 7 the number of groups that have 7 students each in the history session, and select for Remaining students the number of students in the remaining history group, that would be jointly consistent with the given information. Make only two selections, one in each column.
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GMAT Club Olympics 2025 (Day 7): At a summer camp, a class of students 08 Jul 2025, 12:55
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Let the total number of students be N.

For Math group:
one group of 4 and the rest groups of 5
N=4+5x

For Biology group:
one group of 4 and the rest groups of 6
N=4+6y

N must satisfy both conditions that means:
N=4+30k (As LCM of 5 and 6 is 30)

putting k =0,1
k=0: N=4 (Not possible)
k=1: N=34

For History group:
They form groups of 7 and one leftover group with fewer than 7 students.
Thus,
34÷7=4 groups of 7=28 students
remainder=6.


Answer will be:
[*]Groups of 7: 4
[*]Remaining students: 6
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GMAT Club Olympics 2025 (Day 7): At a summer camp, a class of students 08 Jul 2025, 13:33
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let total students be S
from math condition we can conclude that S-4 is divisible by 5
from biology condition we can conclude that S-5 is also divisible by 6
hence together S-4 is divisible by 30
so lets suppose no. of students be 30+4=34
now from history condition all other groups have 7 students each
Remainder(28/7)=4 no of students each in remaining group
no of students in first group which is less than 7=34-28=6
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GMAT Club Olympics 2025 (Day 7): At a summer camp, a class of students 08 Jul 2025, 15:04
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Given:
All students are the same in all 3 sessions (math, biology, history).

Math:
One group of 4,All other groups of 5
⇒ Total = 4 + 5×m

Biology:
One group of 4,All others have 6
⇒ Total = 4 + 6×b

History:
One group has less than 7,All others have 7
⇒ Total = 7×h + r (where r < 7)

We need to find values of h (Groups of 7) and r (Remaining students) such that:

By substituting different values of h and r given in the options, in the above equation
h = 6, r = 4 → Total = 46
Math: 4 + 5×8 = 44
Bio: 4 + 6×7 = 46 satisfies

Groups of 7 = 6
Remaining students = 4
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GMAT Club Olympics 2025 (Day 7): At a summer camp, a class of students 08 Jul 2025, 17:53
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Groups of 7 -> 4
Remaining -> 6

Statements
  • 5x+4
  • 6x+4
  • 7x+(<7)

Before testing notice that 8 and 10 aren't an option for the remaining students.

Groups of 7 options:
2x7=14
4x7=28
6x7=42
8x7=56
10x7=70

+

2 or 4 or 6

If you take 5x+4, you know that the total's last unit has to be 4 or 9 (9,14,19,24,29 etc.)

The options that work are:
4x7+6=34
6x7+2=44
10x7+4=74

The option that also fits 6x+4 (second statement) is 34 = 4x7+6
Bunuel
 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

 



At a summer camp, a class of students participates in three different sessions: math, biology, and history.

• For the math session, the students are divided so that one group has 4 students and all other groups have 5 students each.
• For the biology session, they are divided so that one group has 4 students and all other groups have 6 students each.
• For the history session, the same students are divided so that one group has fewer than 7 students and all other groups have 7 students each.

Select for Groups of 7 the number of groups that have 7 students each in the history session, and select for Remaining students the number of students in the remaining history group, that would be jointly consistent with the given information. Make only two selections, one in each column.
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