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Bunuel
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The students in a school were to be distributed into rows for an assem 04 Jun 2020, 00:53
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25% (medium)

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81% (01:50) correct 19% (02:07) wrong based on 57 sessions

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The students in a school were to be distributed into rows for an assembly. The headmaster earlier thought of distributing the students such that each row had 10 students, but decided against it since this distribution would have led to the last row having only 5 students. Instead, he distributed the students into rows of equal size, with each row having 21 students. Which of the following could be the number of rows in the assembly?

I. 15
II. 30
III. 21

A. I only
B. II only
C. III only
D. I and II only
E. II and III only
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A
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The students in a school were to be distributed into rows for an assem 04 Jun 2020, 02:38
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IMO A
Let the number of students = n
Number of students can be written as 10a +5 or 21b

n = 10 a+ 5 = 5(2a+1) ==> n is odd multiple of 5
Also, n = 21b ==> 3*7*b i.e. 3 & 7 are factors of n

Combining both, n= 2*7*5*p where p is an +ve odd integer

Total number of rows = 2*7*5*p / 21 = 5p (because when students are distirbuted in row so 21, no student is left)
==> # of rows will be odd multpile of 5

From options, only A=15 is valid
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The students in a school were to be distributed into rows for an assem 04 Jun 2020, 07:28
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Bunuel
The students in a school were to be distributed into rows for an assembly. The headmaster earlier thought of distributing the students such that each row had 10 students, but decided against it since this distribution would have led to the last row having only 5 students. Instead, he distributed the students into rows of equal size, with each row having 21 students. Which of the following could be the number of rows in the assembly?

I. 15
II. 30
III. 21

A. I only
B. II only
C. III only
D. I and II only
E. II and III only
Show more


I don't understand how to solve this problem. This is what I got

Given - 10 students per row would leave a remainder of 5 in the last row.
21 students per row equally.

By using the first point "10 students per row and leaving 5 students in the last row" with the answer choices I get the total number of students

A) 15 rows -> (14 rows * 10 students) + (1 row * 5 Students) = 145 students
B) 30 rows -> 290 + 5 = 295 students
C) 21 rows -> 200 + 5 = 205 students

None of these numbers are divisible by 21, which should be the right answer. What am i doing wrong here ?
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gurmukh
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The students in a school were to be distributed into rows for an assem 04 Jun 2020, 10:02
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SiddharthR
Bunuel
The students in a school were to be distributed into rows for an assembly. The headmaster earlier thought of distributing the students such that each row had 10 students, but decided against it since this distribution would have led to the last row having only 5 students. Instead, he distributed the students into rows of equal size, with each row having 21 students. Which of the following could be the number of rows in the assembly?

I. 15
II. 30
III. 21

A. I only
B. II only
C. III only
D. I and II only
E. II and III only


I don't understand how to solve this problem. This is what I got

Given - 10 students per row would leave a remainder of 5 in the last row.
21 students per row equally.

By using the first point "10 students per row and leaving 5 students in the last row" with the answer choices I get the total number of students

A) 15 rows -> (14 rows * 10 students) + (1 row * 5 Students) = 145 students
B) 30 rows -> 290 + 5 = 295 students
C) 21 rows -> 200 + 5 = 205 students

None of these numbers are divisible by 21, which should be the right answer. What am i doing wrong here ?
Show more
Point is number of rows should be such that multiplying it by 21 should be of the form of 10k+5 where k isthe whole number or unit digit of the number is 5. This happens only in option A hence it is the answer.

Posted from my mobile device
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SiddharthR
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The students in a school were to be distributed into rows for an assem 04 Jun 2020, 16:00
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gurmukh
SiddharthR
Bunuel
The students in a school were to be distributed into rows for an assembly. The headmaster earlier thought of distributing the students such that each row had 10 students, but decided against it since this distribution would have led to the last row having only 5 students. Instead, he distributed the students into rows of equal size, with each row having 21 students. Which of the following could be the number of rows in the assembly?

I. 15
II. 30
III. 21

A. I only
B. II only
C. III only
D. I and II only
E. II and III only


I don't understand how to solve this problem. This is what I got

Given - 10 students per row would leave a remainder of 5 in the last row.
21 students per row equally.

By using the first point "10 students per row and leaving 5 students in the last row" with the answer choices I get the total number of students

A) 15 rows -> (14 rows * 10 students) + (1 row * 5 Students) = 145 students
B) 30 rows -> 290 + 5 = 295 students
C) 21 rows -> 200 + 5 = 205 students

None of these numbers are divisible by 21, which should be the right answer. What am i doing wrong here ?
Point is number of rows should be such that multiplying it by 21 should be of the form of 10k+5 where k isthe whole number or unit digit of the number is 5. This happens only in option A hence it is the answer.

Posted from my mobile device
Show more

I'm sorry I still don't get it. Chetan, Bunuel or Karishma, could you please explain :( Question seems pretty straight forward but looks like i'm misunderstanding something
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The students in a school were to be distributed into rows for an assem 07 Jun 2020, 01:48
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SiddharthR
gurmukh
SiddharthR
Bunuel
The students in a school were to be distributed into rows for an assembly. The headmaster earlier thought of distributing the students such that each row had 10 students, but decided against it since this distribution would have led to the last row having only 5 students. Instead, he distributed the students into rows of equal size, with each row having 21 students. Which of the following could be the number of rows in the assembly?

I. 15
II. 30
III. 21

A. I only
B. II only
C. III only
D. I and II only
E. II and III only


I don't understand how to solve this problem. This is what I got

Given - 10 students per row would leave a remainder of 5 in the last row.
21 students per row equally.

By using the first point "10 students per row and leaving 5 students in the last row" with the answer choices I get the total number of students

A) 15 rows -> (14 rows * 10 students) + (1 row * 5 Students) = 145 students
B) 30 rows -> 290 + 5 = 295 students
C) 21 rows -> 200 + 5 = 205 students

None of these numbers are divisible by 21, which should be the right answer. What am i doing wrong here ?
Point is number of rows should be such that multiplying it by 21 should be of the form of 10k+5 where k isthe whole number or unit digit of the number is 5. This happens only in option A hence it is the answer.

Posted from my mobile device

I'm sorry I still don't get it. Chetan, Bunuel or Karishma, could you please explain :( Question seems pretty straight forward but looks like i'm misunderstanding something
Show more

Hello SiddharthR

There are two scenarios given:

1. When each row had 10 students, 5 students are left
2. When each row has 21 students, no student is left

So multiple each option with 21--> this will give you # of students. then divide by 10 and see for which option 5 students are left.

Only option A satisfies both conditions.

Hope this helps.
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The students in a school were to be distributed into rows for an assem 08 Jun 2020, 13:47
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Bunuel
The students in a school were to be distributed into rows for an assembly. The headmaster earlier thought of distributing the students such that each row had 10 students, but decided against it since this distribution would have led to the last row having only 5 students. Instead, he distributed the students into rows of equal size, with each row having 21 students. Which of the following could be the number of rows in the assembly?

I. 15
II. 30
III. 21

A. I only
B. II only
C. III only
D. I and II only
E. II and III only
Show more

Solution:

We see that the number of students is 10x + 5 or 21y for some integers x and y. Notice that x + 1 would be the number of rows if the headmaster distributes the students such that each row has 10 students and y the number of rows if he distributes them such that each row has 21 students. Since the number of students is unique, we must have:

10x + 5 = 21y

Because 10x + 5 has a units digit of 5, we see that y must be an odd multiple of 5 (i.e., 5, 15, 25, etc.) to ensure that 21y also has a units digit of 5. Therefore, of the three values in the given Roman numerals, only 15 can be the number of rows in the assembly.

Answer: A
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The students in a school were to be distributed into rows for an assem 11 Jun 2020, 03:17
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ScottTargetTestPrep
Bunuel
The students in a school were to be distributed into rows for an assembly. The headmaster earlier thought of distributing the students such that each row had 10 students, but decided against it since this distribution would have led to the last row having only 5 students. Instead, he distributed the students into rows of equal size, with each row having 21 students. Which of the following could be the number of rows in the assembly?

I. 15
II. 30
III. 21

A. I only
B. II only
C. III only
D. I and II only
E. II and III only

Solution:

We see that the number of students is 10x + 5 or 21y for some integers x and y. Notice that x + 1 would be the number of rows if the headmaster distributes the students such that each row has 10 students and y the number of rows if he distributes them such that each row has 21 students. Since the number of students is unique, we must have:

10x + 5 = 21y

Because 10x + 5 has a units digit of 5, we see that y must be an odd multiple of 5 (i.e., 5, 15, 25, etc.) to ensure that 21y also has a units digit of 5. Therefore, of the three values in the given Roman numerals, only 15 can be the number of rows in the assembly.

Answer: A
Show more

ScottTargetTestPrep, NitishJain,

If answer is 15 rows then headmaster initially was doing 10*14 +5 = 145 students
Then headmaster did 21*15 =315

So both the cases are giving different number of students, what am I missing ?
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The students in a school were to be distributed into rows for an assem 11 Jun 2020, 03:22
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Bunuel
The students in a school were to be distributed into rows for an assembly. The headmaster earlier thought of distributing the students such that each row had 10 students, but decided against it since this distribution would have led to the last row having only 5 students. Instead, he distributed the students into rows of equal size, with each row having 21 students. Which of the following could be the number of rows in the assembly?

I. 15
II. 30
III. 21

A. I only
B. II only
C. III only
D. I and II only
E. II and III only

Solution:

We see that the number of students is 10x + 5 or 21y for some integers x and y. Notice that x + 1 would be the number of rows if the headmaster distributes the students such that each row has 10 students and y the number of rows if he distributes them such that each row has 21 students. Since the number of students is unique, we must have:

10x + 5 = 21y

Because 10x + 5 has a units digit of 5, we see that y must be an odd multiple of 5 (i.e., 5, 15, 25, etc.) to ensure that 21y also has a units digit of 5. Therefore, of the three values in the given Roman numerals, only 15 can be the number of rows in the assembly.

Answer: A

ScottTargetTestPrep, NitishJain,

If answer is 15 rows then headmaster initially was doing 10*14 +5 = 145 students
Then headmaster did 21*15 =315

So both the cases are giving different number of students, what am I missing ?
Show more

Hello ScottTargetTestPrep,

It’s 21*15= 315

So when divided in rows of 10 students in each row, it will be 315= 31 rows with 10 students each and 5 students left

Posted from my mobile device
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The students in a school were to be distributed into rows for an assem 11 Jun 2020, 04:38
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nitish Jain,
So are you saying answer should be 31 rows and NOT 15 rows?
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The students in a school were to be distributed into rows for an assem 29 Jun 2020, 09:03
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stne
ScottTargetTestPrep
Bunuel
The students in a school were to be distributed into rows for an assembly. The headmaster earlier thought of distributing the students such that each row had 10 students, but decided against it since this distribution would have led to the last row having only 5 students. Instead, he distributed the students into rows of equal size, with each row having 21 students. Which of the following could be the number of rows in the assembly?

I. 15
II. 30
III. 21

A. I only
B. II only
C. III only
D. I and II only
E. II and III only

Solution:

We see that the number of students is 10x + 5 or 21y for some integers x and y. Notice that x + 1 would be the number of rows if the headmaster distributes the students such that each row has 10 students and y the number of rows if he distributes them such that each row has 21 students. Since the number of students is unique, we must have:

10x + 5 = 21y

Because 10x + 5 has a units digit of 5, we see that y must be an odd multiple of 5 (i.e., 5, 15, 25, etc.) to ensure that 21y also has a units digit of 5. Therefore, of the three values in the given Roman numerals, only 15 can be the number of rows in the assembly.

Answer: A

ScottTargetTestPrep, NitishJain,

If answer is 15 rows then headmaster initially was doing 10*14 +5 = 145 students
Then headmaster did 21*15 =315

So both the cases are giving different number of students, what am I missing ?
Show more

Response:

The reason you are getting a different number of students is because you are assuming the number of rows is the same for both scenarios. The number of rows the headmaster initially thought of and the number of rows the headmaster eventually ended up with are actually different. The question is asking for the actual number, not the one the headmaster initially considered.

Since the headmaster did 15 rows with 21 students in each row, there are 21 * 15 = 315 students in total. Since 315 / 10 = 31 R 5, the headmaster initially thought of 32 rows where 31 of the rows had 10 students and the last row had 5 students.
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