At the beginning of the year, the ratio of juniors to seniors in high

Juniors .............. Seniors

3x ........................... 4x (Initially)

After transfers

3x-10 .................. 4x-20

Given that new ratio \(= \frac{3x-10}{4x-20} = \frac{4}{5}\)

x = 30

Seniors initially = 120

Answer = E

Step1: J/S=3/4 => J=3/4(S)
Step 2: (J+10)/(S+20)=4/5
Step 3: substitute Value of J from step1 in step 2 to get s
s=120 (E)

let the total number of juniors and seniors at the beginning of the year be 3x and 4x respectively.

now 10 juniors and 20 seniors are transferred to another school. thus no. of juniors and seniors students left in the school X are 3x-10 and 4x-20 respectively. the ratio of these juniors and seniors students = 4/5
thus we have (3x-10)/(4x-20) = 4/5
15x-50= 16x-80
x=30

thus total no. of seniors at the beginning of the year =4(30)= 120

Here's how I did it:

\(j\) = juniors at beginning of year
\(s\) = seniors at beginning of year

Part 1) \(\frac{j}{s}\) = \(\frac{3}{4}\)

Part 2) \(\frac{(j-10)}{(s-20)} = \frac{4}{5}\)

Reframe Part 1 in terms of j:
\(j\) = \(\frac{3}{4}s\) = \(.75s\)

Substitute j in part 2:

\(\frac{(j-10)}{(s-20)} = \frac{4}{5}\)

= \(\frac{(.75s-10)}{(s-20)} = \frac{4}{5}\)

Now, solve for s, and you get your answer.

I am not sure as to why you added 10, and 20 instead of subtracting it iamevitaerc

Beginning of Year J/S = 3/4 => J = (3/4)S

After transfer : (J-10)/(S-20) = 4/5
= 5J-50 = 4S - 80 => Substitute from the first equation
= (1/4)S = 30
=> S = 120 hence choice D

For these ratio questions I would assume it would not matter which method(two methods listed below) you use for any question throughout the GMAT correct?

Just like this question for example you can use3X/4X = J/S > 3X - 10/4X- 20 = 4/5 to start off the equation and solve as opposed to J/S = 3/4 > J - 10 / S - 20 = 4/5

At the beginning of the year, the ratio of juniors to seniors in high school X was 3 to 4. During the year, 10 juniors and twice as many seniors transferred to another high school, while no new students joined high school X. If, at the end of the year, the ratio of juniors to seniors was 4 to 5, how many seniors were there in high school X at the beginning of the year?

No. Of Juniors = 3x

No. Of Seniors = 4x

New No. Of Juniors = 3x - 10

New No. Of Seniors = 4x - 20

All the above points we have written based on the given information on ratios.

3x - 10/4x - 20 = 4/5

5(3x - 10) = 4(4x - 20)

15x - 50 = 16x - 80

x = 30

No. Of Seniors at the beginning = 4x = 4 * 30 = 120

Hence, the Answer is E

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Originally -

\(\frac{Juniors}{Seniors} = \frac{3x}{4x}\)

\(\frac{3x - 10}{4x - 20} = \frac{4}{5}\)

\(15x - 50 = 16x - 80\)

\(x = 30\)

So, The total number of seniors in high school is \(4*30 = 120\)

Thus, the correct answer must be (E) 120
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At the beginning of the year, the ratio of juniors to seniors in high school X was 3 to 4. During the year, 10 juniors and twice as many seniors transferred to another high school, while no new students joined high school X. If, at the end of the year, the ratio of juniors to seniors was 4 to 5, how many seniors were there in high school X at the beginning of the year?

A. 80
B. 90
C. 100
D. 110
E. 120

Suppose, initially the ratio was: 3x/4x
Due to transfer, it stands: 3x-10/4x-20 = 4/5
upon calculation: x = 30
the seniors were: 4*30 = 120

Answer: E

There is a time-saving trick to do such questions:
look at the numbers in the ratios (here the numbers are 3,4 and 5)
now analyze the answer choices and see which number is a multiple of all 3 numbers in the ratio.
option E satisfies the condition, so it is the correct choice.

Disclaimer: I don't know if this is a legit trick but I tried this trick on a few similar questions and it worked for me.
please correct me if I'm wrong.

Thx in advance!

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