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kawal27
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The ratio of the number of students in two classrooms, C1 and C2, is 2 07 Sep 2020, 02:01
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The ratio of the number of students in two classrooms, C1 and C2, is 2 : 3. It is observed that after shifting 10 students from C1 to C2, the ratio is 3 : 7. Further, how many students have to be shifted from C2 to C1 for the new ratio to become 9 : 11?


A)10

B)15

C)20

D)25

E)30
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B
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rajatchopra1994
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The ratio of the number of students in two classrooms, C1 and C2, is 2 07 Sep 2020, 02:12
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Explanation:

Consider the number of students in C1 and C2 are 2x and 3x
Given (2x−10):(3x+10)=3:7
14x-70 = 9x+30
5x=100
x=20 --- 1
Let y be the number of students to be shifted from C2 to C1.
Substitute value of x in (2x−10+y):(3x+10−y)=9:11
(30+y):(70-y)=9:11
330+11y = 630-9y
20y = 300
y= 15

IMO-B
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The ratio of the number of students in two classrooms, C1 and C2, is 2 07 Sep 2020, 09:01
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kawal27
The ratio of the number of students in two classrooms, C1 and C2, is 2 : 3. It is observed that after shifting 10 students from C1 to C2, the ratio is 3 : 7. Further, how many students have to be shifted from C2 to C1 for the new ratio to become 9 : 11?
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Making both ratios out of the same total (of ten) so we can properly compare them, the initial ratio is 4 to 6. After moving ten students, the ratio becomes 3 to 7. Since a change of one in the ratio corresponds to an actual change of ten, all of our numbers must be ten times those in the ratios, so we started with 40 and 60 students. Moving ten of them leads us to 30 and 70 students. If we're going to now move students from the second class to the first to get to a 45 to 55 ratio, we'd need to move 15 students.
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The ratio of the number of students in two classrooms, C1 and C2, is 2 07 Sep 2020, 13:01
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rajatchopra1994
Explanation:

Consider the number of students in C1 and C2 are 2x and 3x
Given (2x−10):(3x+10)=3:7
14x-70 = 9x+30
5x=100
x=20 --- 1
Let y be the number of students to be shifted from C2 to C1.
Substitute value of x in (2x−10+y):(3x+10−y)=9:11
(30+y):(70-y)=9:11
330+11y = 630-9y
20y = 300
y= 15

IMO-B
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So every ratio can be expressed as a fraction, e.g.:

3:4 = \(\frac{3}{4}\)

Is that correct? If yes, how could the question be solved, if there were three ratios involved?
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The ratio of the number of students in two classrooms, C1 and C2, is 2 28 Dec 2020, 04:27
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Consider the number of students in C1 and C2 are 2x and 3x
Given (2x−10):(3x+10)=3:7
14x-70 = 9x+30
5x=100
x=20 __ (1)

Let y be the number of students to be shifted from C2 to C1.
Substitute value of x in (2x−10+y):(3x+10−y)=9:11
(30+y):(70-y)=9:11
330+11y = 630-9y
20y = 300
y=15
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The ratio of the number of students in two classrooms, C1 and C2, is 2 28 Dec 2020, 05:19
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kawal27
The ratio of the number of students in two classrooms, C1 and C2, is 2 : 3. It is observed that after shifting 10 students from C1 to C2, the ratio is 3 : 7. Further, how many students have to be shifted from C2 to C1 for the new ratio to become 9 : 11?


A)10

B)15

C)20

D)25

E)30
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Since the resulting ratio is 9:11 -- and the parts of this ratio have a sum of 20 -- the total number of students must be a multiple of 20.
Test whether a multiplier of 20 for the initial 2:3 ratio satisfies the following condition in the prompt:
Shifting 10 students from C1 to C2, the ratio is 3 : 7.

\(\frac{C1}{C2} = \frac{2}{3} = \frac{2*20}{3*20} = \frac{40}{60}\)
If 10 students transfer from C1 to C2, we get:
\(\frac{C1}{C2}=\frac{40-10}{60+10 }=\frac{30}{70}=\frac{3}{7}\)
Success!

How many students have to be shifted from C2 to C1 for the new ratio to become 9:11?
When the correct answer is subtracted from the 70 students currently in C2, the result must be a multiple of 11.
Only B is viable.
If 15 students transfer from C2 to C1, we get:
\(\frac{C1}{C2}=\frac{30+15}{70-15} = \frac{45}{55} = \frac{9}{11}\\
\)
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The ratio of the number of students in two classrooms, C1 and C2, is 2 04 Feb 2026, 09:11
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Deconstructing the Question
Given
C1:C2 = \(2:3\)
After moving 10 students from C1 to C2, ratio becomes \(3:7\).
Find how many students x must be moved from C2 to C1 so the ratio becomes \(9:11\).

Step-by-step
Let C1 = \(2k\), C2 = \(3k\).

After moving 10 from C1 to C2:
C1 = \(2k - 10\), C2 = \(3k + 10\).

Set the ratio:
\(\frac{2k-10}{3k+10}=\frac{3}{7}\)

Cross-multiply:
\(7(2k-10)=3(3k+10)\)
\(14k-70=9k+30\)
\(5k=100 \Rightarrow k=20\)

So after the first move:
C1 = \(40-10=30\), C2 = \(60+10=70\).

Now move x from C2 to C1:
C1 = \(30+x\), C2 = \(70-x\).

Target ratio:
\(\frac{30+x}{70-x}=\frac{9}{11}\)

Solve:
\(11(30+x)=9(70-x)\)
\(330+11x=630-9x\)
\(20x=300 \Rightarrow x=15\)

Answer: 15
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The ratio of the number of students in two classrooms, C1 and C2, is 2 22 Feb 2026, 14:21
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Hi Ian, i have doubt about this way os calculatoin.

If you have multiplied by 2 the first part ( 2:3 -> 4:6), why you dind ́t do it in the second ( 3:7 -> 6:14)? The second change for me would make sense considering that we want to achieve the 9:11, which leads to 20. But if I did this, i would get 30 as a result, not 15. My main question here, is how can I understand when to stop multipliyng and start doing the division to get to the result.

Tnks :)
IanStewart


Making both ratios out of the same total (of ten) so we can properly compare them, the initial ratio is 4 to 6. After moving ten students, the ratio becomes 3 to 7. Since a change of one in the ratio corresponds to an actual change of ten, all of our numbers must be ten times those in the ratios, so we started with 40 and 60 students. Moving ten of them leads us to 30 and 70 students. If we're going to now move students from the second class to the first to get to a 45 to 55 ratio, we'd need to move 15 students.
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