The ratio of the number of students in two classrooms, C1 and C2, is 2

Question

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

A)10

B)15

C)20

D)25

E)30

rajatchopra1994 · Accepted Answer

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

IanStewart · Answer

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.

ceanma · Answer

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?

HarshBavishi · Answer

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

GMATGuruNY · Answer

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}
 
 B

luisdicampo · Answer

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

luisaacsiqueira · Answer

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

ronakpujari94 · Answer

He did it in the first part because he wanted to find out the total number of students. Once in the first part we know the total number of students multiplying in the second part serves no purpose as the number of students remain same. There were 40,60 students in the beginning, after the first movement C1 -> C2 there were 30,70; now the question asks after another movement of C2 ->C1, how would we get the ratio of 9:11 for the same 100 students i.e 45,55 students in C1 and C2.  Answer :- by moving 15 students from C2 to C1

The ratio of the number of students in two classrooms, C1 and C2, is 2

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The ratio of the number of students in two classrooms, C1 and C2, is 2

Prep Toolkit

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