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GMAT Club Olympics 2025 (Day 7): At a summer camp, a class of students

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MinhChau789

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09 Jul 2025, 04:38

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From the info, N must be in the form of:
(1) N = 4+ 5a . So N = 9, 14, 19, 24, 29, 34,...
(2) N = 4 + 6b . So N = 10, 16, 22, 28, 34,...
(3) N = 7c + r (where r <7)

From (1) & (2), we know that N can be 34. So N must be in form of 30a + 34. From the options, only "Groups of 7" = 4, and remaining students = 6 can satisfy.

LastHero

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09 Jul 2025, 04:43

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math session: (students-4) is a multiple of 5
biology session: (students-4) is a multiple of 6

(students-4) can be 30, 60, 90, 120 and so on
students can be 34, 64, 94, 124 and so on

Check the numbers in the answers:
students=7*(Groups of 7) + (Remaining students)
34=7*(Groups of 7) + (Remaining students)
34 = 7*4 + 6

The right answers:
Groups of 7=4
Remaining students=6

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09 Jul 2025, 04:53

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Bunuel

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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.

Let total student be T

For Maths
T = 4 + 5a

For Biology
T = 4 + 6b

For History
T < 7 + 7c

Now
4 + 5a = 4 + 6b
=> 5a = 6b
=> 5a and 6b could be 30. 60, 90, 120 .......

Hence T = 34, 64, 94, 124 ........

Let's take T = 34 = 7c + k ---- [c is groups of 7 and k is remaining students]
=> c = 4 and k = 6
this combination is available in the answer choices

Let's try T = 64 = 7c + k
=> c = 9 and k = 1

Let's try T = 94 = 7c + k
=> c = 13 and k = 3

Hence

Groups of 7 = 4
Remaining student = 6

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09 Jul 2025, 04:58

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math = 4 + 5a = total -> total-4 must be a multiple of 5
biology = 4 + 6b = total -> total-4 must be a multiple of 6
history = c + 7d = total, with c<7

if total-4 must be a multiple of 5 and 6, total can be: 34, 64, 94...

if c+7d=34, c=6 and d=4, that are in the options

Groups of 7 = 4 and Remaining students = 6

ODST117

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09 Jul 2025, 05:11

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Total number of students in the math session -> $4+5x_1$ ($x_1$ is the number of groups with $5$ students in them)
Total number of students in the biology session -> $4+6x_2$ ($x_2$ is the number of groups with $6$ students in them)

The total number of students doesn't change per session, therefore, students in the math session = students in the biology session

$4+5x_1=4+6x_2$

$5x_1=6x_2$

The minimum value of LHS and RHS where they are equal is $30$, when $x_1=6$ and $x_2=5$

The total number of students $= 30+4 = 34$

The history session had groups with $7$ students each and one group with less than $7$ students.
The max number of groups of $7$ students that can be made with $34$ students is $4$ (because $7*4 = 28$. If we said 5 groups, that would mean $7*5=35$ students total, which is more than what we have)
If $28$ students are in $4$ groups of $7$, that leaves $6$ students in the group which has less than $7$ students.

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09 Jul 2025, 05:48

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Correct Answer: Group of 7 students= 4 and Remaining students= 6
For this question check by option method,
Let’s check with options and try to find total students:
2*7= 14
4*7= 28
6*7= 42
8*7= 56
10*7= 70

Now by random take group= 4 and use remaining= 6
Now total= 7*4+6= 28+6= 34

• For the math session, the students are divided so that one group has 4 students and all other groups have 5 students each.
If we divide 5 students each so here we have 6 groups and 4 are the remaining students. Hence we got this correct.

• For the biology session, they are divided so that one group has 4 students and all other groups have 6 students each.
Similarly if we divide 6 students each so here we have 5 groups and 4 are the remaining students. Hence we got this also correct.

• 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.
Lastly if we divide 7 students each so here we have 4 groups and 6 are the remaining students as this less than 7. Hence we got this also correct.

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09 Jul 2025, 05:52

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Based on the Math and Biology groups whose numbers are given with 4 being constant we need to find the LCM of 5 and 6 to find a number that would easily fit the groupings without a remainder.
The LCM of 5 and 6 is 30 meaning those distributed in groups of 5 and 6 could be 30 or multiples of 30 meaning total students are either 34, 64, 94......
Testing those values with the history class distribution the only available answer from the choices is
4 Groups of 7
6 Remaining students meaning the total students are 34

Bunuel

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09 Jul 2025, 05:58

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Total no. of students = N
For maths, 1 group has 4 , remaining have 5
-> N – 4 is a multiple of 5.
For biology, 1 group has 4, remaining have 6
-> N – 4 is also a multiple of 6.

LCM (5, 6) = 30
-> N – 4 = 30
-> N = 34

For history, we can split 34 students into groups of 7, which will give us 4 full groups and 6 remaining students

Answer :
Groups of 7: 4
Remaining students: 6

IIIJOHNNYSIII

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Let math, biology, and history be represented by m, b, h.
m has 1 team with 4 members, rest is split into teams of 5 members each
b has 1 team with 4 members, rest is split into teams of 6 members each
n has 1 team with < 7 members, rest is split into teams of 7 members each

From first two statements, taking LCM of 5 & 6 we have 30+4, say 34 students in total.
Considering the last statement 7*4 = 28 and the rest in 1 team which has 6 members which is < 7.
That is a total of 34 students and all the conditions are satisfied and the values are consistent.

Correct answers are,
Groups of 7 : 4
Remaining students : 6

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09 Jul 2025, 06:09

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From the math session, N = 4 + 5x (where x is the number of groups with 5 students) => N will end in either 4 or 9 since 5X will end in (0,5)
From the biology session, N = 4 + 6y (where y is the number of groups with 6 students) => N will be even
=> N will end in "4"
This also gives that (N-4) is multiple of 30 LCM of (5,6)
From the history session, N = 7G + R we know that R<7 so R can be 2,4,6

Try options, start with B, G=4 ,

If R = 2, N = 7(4) + 2 = 28 + 2 = 30 (Ends in 0, not 4)
If R = 4, N = 7(4) + 4 = 28 + 4 = 32 (Ends in 2, not 4)
If R = 6, N = 7(4) + 6 = 28 + 6 = 34 (ends with 4 and (n-4) is multiple of 30.

So answer is 4 and 6.

Bunuel

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09 Jul 2025, 06:51

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Total number of students in

Maths = 5k+4

Biology= 6m+4

The plausible number for the total number of students is a multiple of 5 and 6
which is 30. Thus N=34

if 34 is divided into groups of 7 students. How many groups would we have

34/7= 4 R6

So we would have 4 groups of 7 Students with the remaining 6 put into the last group

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Let M = 4*1 + 5n
B = 4 + 6m
H = 7p + (<7)

We need to find the split of H i.e. History session
M = B
5n = 6m
Presume n = 6 and m=5 and total = 34

We can see that:
7*4+6 = 34

Groups of 7: 4
Remaining students: 6

Bunuel

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09 Jul 2025, 07:07

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Three different sessions: math, biology, and history.

Let T be the total number of students

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

=> T = 5a + 4 (a>=1)

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

=> T = 6b + 4 (b>=1)

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 = 7c + d (d < 7 and c>=1)

Now we have 2 equations,
T = 5a + 4
T = 6b + 4

=> T ≡4 mod 5 and T ≡4 mod 6
=> T - 4 must be divided by LCM(5,6) = 30
Thus,
T = 30k + 4 for an integer k

From equation 3, and above condition we need to find k where T when divided by 7 gives remainder (d) < 7
Lets try k = 1.
T = 30*1 + 4 = 34
34 divided by 7 gives a remainder 6

=> Groups of 7 = 4 and remainder students = 6
If we were to tally this with the other equations,
For Maths ; 5a + 4 = 34 => a = 6
For Biology; 6b + 4 = 34 => b = 5
Both are integers and satisfy above a and b constraint so its valid

Groups of 7 = 4
Remaining students = 6

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09 Jul 2025, 07:18

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Gievn:
Math = 5x+4 (x = no. of groups with 4 people)
Biology = 6y+4 (y = no. of groups with 6 people)
History: 7z + (>7) (z = no. of groups with 7 people)
Determine the total students in the summer camp
Consider math and biology: Total student has to be LCM(5,6) + 4
Possible values: 4,34,64...
Now, the same total should be written in the form 7z+ >7
If total = 34, 7*4+6
Matches the criteria and also there in the options. Test the next value
If total = 64, 7*9 + 1
Both 9 and 1 are not available in Option choices.
Answer 4 and 6

Bunuel

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09 Jul 2025, 07:25

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Given:
All students divided
For maths groups of 5 students with one group of 4 (this means no. of group is multiple of 5 and remainder is 4)
For Biology, Groups of 6 students with one group of 4 (No. of groups is multiple of 6 with remainder 4)
History, (7 students per group and remainder is unknown)

So we need to first see how many students are there in class.

So if we say 6 groups of 5, 5groups of 6 is created then 34 is total number of students in class.

So no. of students when divided by 34/7, 4 total groups with remainder 6 students.

and this answer is available as well.

Answer: 4 groups of 7 students and 6 students in one group (remainder).

Bunuel

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09 Jul 2025, 07:39

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Math: 4 + 5M | Bio: 4 + 6B | History: <7 + 7H
- From math & bio, we have the ratio M:B = 6:5 and we can plug back in to find the common total number of students: 34, 64, 94..
- To satisfy the history class w/ the above total, we only have H (groups of 7) = 4 & its remaining students = 6

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09 Jul 2025, 07:48

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Maths:5x+4
biology:6y+4
history:7z+a (a<7)
Let the total number of students be N.
So,
4+5x=4+6y
=>5x=6y
=>x/y=6/5
y should be a multiple of 5
N=4+6×5=34
For history it can be 4*7+6 (since we know a is less than 7)
Groups of 7: 4
Remaining students: 6

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09 Jul 2025, 07:54

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Maths = 4 + 5x
Biology = 4 + 6y

As students are same in both subjects we can equate both equation

x/y = 6/5, Taking x = 6 and y = 5 give the total students = 30 + 4 = 34. Which satisfy the given constraints according to given options.

34/7 = 4 groups of 7 plus 6 students in History.

Groups of 7 = 4 and Remaining students = 6

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