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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.

Math session has 4+5k students which can be 9,14,19,24,29,34,39,44,49 likewise
Biology session has 4+6k which can be 10,16,22,28,34,40,46,52 likewise

Math session & Biology session can have 30K+4 students can be 34,64,94 likewise

Lets total students be 34 can be 7k+x or 4 groups of 7 & 6 students of remaining group

64 students of 9 groups of 7 and 1 group of 1 students

94 students 13 groups of 7 and one group of 3 students

from option it can be seen total students are 34 thus 4 groups of 7 and remaining students 6
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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.
If we simplify the question, It comes to that the number of students as S

So S=5x+4 = 6y+4. This means S can be 34,64,94 etc..

We can now look for the corresponding pair in the options:

34 = 7*4 + 6.
64 = 7*9+1

As we don't have (9,1) pair. We have (4,6) pair .

So there are IMO 4 groups of 7 students, and 6 are remaining.

Groups of 7 : 4
Remaining students : 6
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In this we can do easily by LCM.
For maths one group is having 4 and other groups are having 5 so we can take as 4 5 5 5 5 5 5=34
For biology 4 6 6 6 6 6=34
For history some groups have 7 and one has less than that. Let us see how many can we make- 7 7 7 7 and 6=34.
So 4 groups have 7 and one group has 6 which is less than 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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we are looking for solution that satisfy the constrains:

5k1 + 4 = total student ... (math constr.)
6k2 + 4 = total student ... (bio contr.)
7k3 + n = total student .. where (n < 7) ... (history constrain)

I did brute force on the available choices, but you can take a smart approach of the anything divisble by 5, 6; leaving remainder 4;

7 x (2 ) + (6) = 34

2 groups of 7 students each and 6 remaining students.
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Let the total number of students be S.

From the math session: one group has 4 students, others have 5. so S = 5x + 4

From the biology session: one group has 4, others have 6 → so S = 6y + 4.

Try small values satisfying both: test S=34 works for both math and biology patterns.

For history, 34÷7 = 4 groups of 7 and 6 students remaining.

Answer: Groups of 7 = 4, Remaining students = 6.
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Math & biology session have groups of 5 & 6 students each plus 4 students. So the total no of students must be multiple of LCM of 5 & 6 plus 4

\(LCM (5,6)=30\)

Total students \(T=30k+4\)

Groups of 7:

when \(k=1\), \(T=34\), max multiple of 7 within 34 is 28, which can be divided into 4 groups of 7 each. Remaining one group will have 6 students

when \(k=2\), \(T=64\), max multiple of 7 within 64 is 63, which can be divided into 9 groups of 7 each. Remaining one group will have 1 students

Looking at the options, we have both 4 & 6 as choices

Groups of 7: B

Remaining students: C
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the group of students can be divided into:
5m + 4........(1)
or
6b + 4.........(2)
or
7h + x.........(3)

equating 1 and 2
we get:
5m = 6b
for m = 6 and b = 5 we get the total students as 34

so 7h + x = 34
hence h = 4, x=6

so we get the answer
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For math session, let number of groups with 5 students each = m
Then number of students = 5m+4

For biology session, let number of groups with 6 students each = b
Then number of students = 6b + 4

The number of students are the same in both cases:
5m + 4 = 6b + 4
5m = 6b

since 5 and 6 have no common divisors except 1, minimum value of m must be 6 and minimum value of b must be 5 such that
5*6 = 6*5
Soo there are 6 groups of 5 + 1 group of 4 in math session, and 5 groups of 6 + 1 group of 4 in biology session

So total number of students = 5m+4 = 30+4 = 34

Now for the history session, dividing 34 by 7 gives quotient 4 and remainder 6
Hence there must be 4 groups of 7 and remaining students must be 6 (<7) in the history session
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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.
We are given: A group of students participates in 3 sessions (math, biology, and history), and the total number of students is the same in all sessions (Suppose it's n).
We are told how they are grouped in each session:
- Math session: n = 5a + 4 (1)
- Biology session: n = 6b + 4 (2)
- History session: n = 7c + r (r<7) (3)
From (1) and (2) => n-4 = 30k (= 30, 60, ...) => n = 34, 64, ....
If n = 34, from (3) we have: c = 4 and r = 6
If n = 64, from (3) we have: c = 9 and r = 1
=> Final question: Groups of 7 = 4
Remaining students = 6
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We know that the number of total students is such that it leaves remainder 4 for both groups of 5 and groups of 6.
That means the total number of students - 4 would result in a number divisible by both 5 and 6.
Taking LCM 30
Total number would be 34.
Applying groups of 7
maximum number of groups possible would be 4 (7*4 = 28)
And the remaining group will have 6 students.
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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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given 5y+4 = 6x+4
5y=6x
The values of 5y and 6x could be 30, 60 etc
if it is 30+4 = 34 then
there could be 4 groups of 7 and 6 remaining students
if it was 64 then there would be 9 groups with 1 leftover student which is not in the options.
Hence there are 4 groups with 6 remaining students
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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Bunuel
 


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

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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.
For Math session -- > Total students = 5x+4
Biology session -- > Total students = 6y+4
History session -- > Total students = 7z+ (less than 7 students )

Since, all above equations represent the total students :-

5x+4 = 6y+4
x=6y/5 ............

x is an integer hence y is a multiple of 5.
y = 5, 10, 15, 20..........5k
x= 6, 12, 18, 24...........6k

Total Students : Maths => 34, 64, 94, ..........
: Biology => 34, 64, 94...........

History Session :
for Total Students = 34 = 7*4 + 6

Number of groups with 7 students = 4, Remaining Students = 6
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If the remainder is 4, the number minus 4 is divisible by 5 (math) and divisible by 6 (biology).
Multiples of 5 and 6: 30, 60...

Total number of students can be 34, 64...

(Groups of 7)*7 + Remaining students = 34

4*7 + 6 = 34

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.
Groups of 7 Remaining students
2
4
6
8
10

M: 5m+4
B: 6b+4
H: 7h+(<7)

5m+4=6b+4
Thus, m is a multiple of 6 and b is a multiple of 5
Thus if we try first value of M or B= 34

now 7h+(<7)=34 is possible if h=4 and in that case remainder is 6

Ans B, C
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math: students can be 9, 14, 19, 24, 29, 34, 39, 44, 49, 54, 59, 64...
biology: students can be 10, 16, 22, 28, 34, 40, 46, 52, 58, 64...

34 and 64 are in the two lists.

7*g + r = 34
True if g=4 and r=6, that are in the possible answers

Correct answers: Groups of 7 = 4 and Remaining students = 6
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M: 4+5m ---- (1)
B: 4+6b ---- (2)
H:<7+7h ----- (3)

From 1 & 2, total is 30.

30 = r + 7h, this would match to answer of 6 + 7 *4

Answer: 4, 6
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