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COolguy101
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Among 20 separate classes, there are either 27 or 28 students in each 01 Nov 2021, 01:47
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Question Stats:

59% (03:01) correct 41% (02:46) wrong based on 162 sessions

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Among 20 separate classes, there are either 27 or 28 students in each class. The school now plans to open another separate class such that the average number of students of all the 21 classes is 0.5 less than the average of the original 20 classes. How many students are in the new class?

A. 10
B. 15
C. 16
D. 17
E. 18
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D
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Among 20 separate classes, there are either 27 or 28 students in each 01 Nov 2021, 04:15
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COolguy101
Among 20 separate classes, there are either 27 or 28 students in each class. The school now plans to open another separate class such that the average number of students of all the 21 classes is 0.5 less than the average of the original 20 classes. How many students are in the new class?

A. 10
B. 15
C. 16
D. 17
E. 18
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Let the average of 20 classes be x. As, the strength of each class is 27 or 28, x should be between 27 and 28, both inclusive.

The average of 21 classes is x-0.5, so the total students now = 21(x-0.5) = 21x -10.5

The strength of the new class = Final total - Initial total = 21x-10.5 - (20x) = x-10.5

We know that \(27\leq x\leq 28\), so subtracting 10.5 from each, we get \(27-10.5\leq x-10.5\leq 28-10.5\) or \(16.5 \leq x-10.5\leq 17.5\)
But x-10.5 HAS to be a positive integer, so only possible value in the range is 17.


D
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Among 20 separate classes, there are either 27 or 28 students in each 01 Nov 2021, 11:38
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COolguy101
Among 20 separate classes, there are either 27 or 28 students in each class. The school now plans to open another separate class such that the average number of students of all the 21 classes is 0.5 less than the average of the original 20 classes. How many students are in the new class?

A. 10
B. 15
C. 16
D. 17
E. 18
Show more
Hello, everyone. I like the solution that chetan2u has provided above. For the less algebraically inclined puzzle-solver, it is easy to use numbers for an open-ended average question such as this one. If there are either 27 or 28 students per class, and there are 20 separate classes, then nothing prohibits us from splitting the difference for ease of calculation: 10 classes of 27 students and 10 of 28. As with many average (arithmetic mean) questions, we are looking for a sum.

\(10 * 27 = 270\)

\(10 * 28 = 280\)

\(270 + 280 = 550\)

We can set the number of students to 550. The actual number does not matter, since all we are looking to do is reduce the average number of students per class by 0.5 once this twenty-first class is added. Although we can deduce that the average number of students per class in this scenario will be split down the middle at 27.5, you can also figure out the current average if it makes you feel more comfortable.

\(\frac{550}{20}=\frac{55}{2}=27.5\)

Next, we have to reduce this average and multiply by 21 (classes) to get the new sum.

\(27.5 - 0.5 = 27\)

\(27 * 21 = 567\)

All we need to do now is find the difference in the number of students between the new sum and the old sum.

\(567 - 550 = 17\)

The answer must be (D). I hope this helps someone who may be hesitant to tackle the question with numbers, thinking that the exact value for the original number of students is somehow vital to solving the question. (Picking other numbers upfront would lead to decimal answers that were still closest to 17, rounding up or down as necessary.)

Good luck with your studies.

- Andrew
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Among 20 separate classes, there are either 27 or 28 students in each 01 Nov 2021, 22:53
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COolguy101
Among 20 separate classes, there are either 27 or 28 students in each class. The school now plans to open another separate class such that the average number of students of all the 21 classes is 0.5 less than the average of the original 20 classes. How many students are in the new class?

A. 10
B. 15
C. 16
D. 17
E. 18
Hello, everyone. I like the solution that chetan2u has provided above. For the less algebraically inclined puzzle-solver, it is easy to use numbers for an open-ended average question such as this one. If there are either 27 or 28 students per class, and there are 20 separate classes, then nothing prohibits us from splitting the difference for ease of calculation: 10 classes of 27 students and 10 of 28. As with many average (arithmetic mean) questions, we are looking for a sum.

\(10 * 27 = 270\)

\(10 * 28 = 280\)

\(270 + 280 = 550\)

We can set the number of students to 550. The actual number does not matter, since all we are looking to do is reduce the average number of students per class by 0.5 once this twenty-first class is added. Although we can deduce that the average number of students per class in this scenario will be split down the middle at 27.5, you can also figure out the current average if it makes you feel more comfortable.

\(\frac{550}{20}=\frac{55}{2}=27.5\)

Next, we have to reduce this average and multiply by 21 (classes) to get the new sum.

\(27.5 - 0.5 = 27\)

\(27 * 21 = 567\)

All we need to do now is find the difference in the number of students between the new sum and the old sum.

\(567 - 550 = 17\)

The answer must be (D). I hope this helps someone who may be hesitant to tackle the question with numbers, thinking that the exact value for the original number of students is somehow vital to solving the question. (Picking other numbers upfront would lead to decimal answers that were still closest to 17, rounding up or down as necessary.)

Good luck with your studies.

- Andrew
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But sir, I make
total sum/20 = sum/21 -0.5
After solving , sum = 210 and which have to divide into 21 classes and I got 10 as average. What I am making mistakes here Sir?
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Among 20 separate classes, there are either 27 or 28 students in each 02 Nov 2021, 08:26
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COolguy101

But sir, I make
total sum/20 = sum/21 -0.5
After solving , sum = 210 and which have to divide into 21 classes and I got 10 as average. What I am making mistakes here Sir?
Show more
Hello, COolguy101. I was not saying that anything is wrong with figuring out the exact numbers here, just that doing so is not strictly necessary. The problem merely stipulates that among 20 separate classes, there are either 27 or 28 students in each class. We could carve up the number of students any way we wanted and still arrive at the correct answer. For instance, say there were 19 classes with 27 students and a single class with 28 students.

\(\frac{19 * 27 + 1 * 28}{20} = 27.05\)

The average would be 27.05 students per class. We could subtract 0.5 from this average the same as we could for any other average.

\(27.05 - 0.5 = 26.55\)

If we multiplied this new average by 21 classes, we would derive the new sum.

\(26.55 * 21 = 557.55\)

Subtracting the original number of students from the new sum, we would still get the same answer.

\(557.55 - 541 = 16.55\)

16.55 rounds to 17, so the answer is (D), all the same. Had we approached the problem from the upper extreme (say, 1 class with 27 students and 19 classes with 28 students), we would still get a number that would round to 17. For many test-takers, sometimes it can help to let go of the inner fact-checker and use logic to work more efficiently toward an answer. That was the point I was hoping to make, nothing more.

Thank you for following up.

- Andrew
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Among 20 separate classes, there are either 27 or 28 students in each 02 Nov 2021, 09:45
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COolguy101

But sir, I make
total sum/20 = sum/21 -0.5
After solving , sum = 210 and which have to divide into 21 classes and I got 10 as average. What I am making mistakes here Sir?
Hello, COolguy101. I was not saying that anything is wrong with figuring out the exact numbers here, just that doing so is not strictly necessary. The problem merely stipulates that among 20 separate classes, there are either 27 or 28 students in each class. We could carve up the number of students any way we wanted and still arrive at the correct answer. For instance, say there were 19 classes with 27 students and a single class with 28 students.

\(\frac{19 * 27 + 1 * 28}{20} = 27.05\)

The average would be 27.05 students per class. We could subtract 0.5 from this average the same as we could for any other average.

\(27.05 - 0.5 = 26.55\)

If we multiplied this new average by 21 classes, we would derive the new sum.

\(26.55 * 21 = 557.55\)

Subtracting the original number of students from the new sum, we would still get the same answer.

\(557.55 - 541 = 16.55\)

16.55 rounds to 17, so the answer is (D), all the same. Had we approached the problem from the upper extreme (say, 1 class with 27 students and 19 classes with 28 students), we would still get a number that would round to 17. For many test-takers, sometimes it can help to let go of the inner fact-checker and use logic to work more efficiently toward an answer. That was the point I was hoping to make, nothing more.

Thank you for following up.

- Andrew
Show more

Okay, I got your point, But if I do this,
\(\frac{19 * 28 + 1 * 27}{20} = 27.95\) and subtracting 0.5 gives \(27.05 - 0.5 = 27.45\), multiplying this with 21, gives 576.45.
subtracting new sum from old one,
\( 559- 576.45 = 17.45\), this time I got 17.45, which is roughly equal to 18 , and 18 is also present in answer choices. Then, how to make a good selection in this case?
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Among 20 separate classes, there are either 27 or 28 students in each 02 Nov 2021, 12:01
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COolguy101

Okay, I got your point, But if I do this,
\(\frac{19 * 28 + 1 * 27}{20} = 27.95\) and subtracting 0.5 gives \(27.05 - 0.5 = 27.45\), multiplying this with 21, gives 576.45.
subtracting new sum from old one,
\( 559- 576.45 = 17.45\), this time I got 17.45, which is roughly equal to 18 , and 18 is also present in answer choices. Then, how to make a good selection in this case?
Show more
No, 17.45 would round down to 17, so the answer would still be (D). The only way you could get to 18 by rounding is by ignoring the implied information that at least one class contains 27 students. (Otherwise, why would the problem stipulate either 27 or 28 students? It would presumably say that each class had the same number of students.) Such an approach, starting with 28 students in each of 20 classes, would lead to a difference of exactly 17.5, and that would, in fact, round up. However, when you consider that of all the potential combinations of 27/28 students, from a split of 20/0 to 0/20, a single one would lead to 17.5—i.e. each other combination would lead to a rounded value of 17—a strong case can be made for 17, rather than for the outlier in 18.

I apologize for the confusion. I was hoping to point out a way to make things simpler, but ultimately, whichever way you feel comfortable solving a problem is the way you should go about solving it, as long as it leads to the correct answer.

- Andrew
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Among 20 separate classes, there are either 27 or 28 students in each 02 Nov 2021, 12:35
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Given: Among 20 separate classes, there are either 27 or 28 students in each class. The school now plans to open another separate class such that the average number of students of all the 21 classes is 0.5 less than the average of the original 20 classes.

Asked: How many students are in the new class?

Let the Average of 20 classes be A and number of students in the new class be x.

20A + x = 21(A-.5) = 21A - 10.5
A = x + 10.5

A is between 27 & 28
Let A = 27.y

27.y = x + 10.5
x = 27.y - 10.5 = 17.(y-5)

Since x is number of students in a class, it is an integer

x = 17

IMO D
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Among 20 separate classes, there are either 27 or 28 students in each 05 May 2025, 03:20
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Let the no. class with 27 student be "X" and with 28 students be "20-X" and new students be K

0.5 = [27X + 28(20-x)]/20 - [27X + 28(20-x) + K]/21

on solving
X = 350 -20K

Only value of k for which X lies b/w 1 to 20 is 17

so, D
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