How many students are there in Class A?

[eddy2017]

There are no other tutor present at the forum right now - I'll get in the foray just to be expedient.

You have 3 "linear" equations - with 3 unknowns.

The general rule of

"solving for 'n' unknowns" from 'n' linear equations is - we systematically reduce the equations, such that we are left with "one equation and 1 unknown".

In this case we plan to reduce the equations - first two equations and two unknowns.

There are several methods (which you may learn later) to reduce the number of equations and variables.

In this case simple observation tells me that if we subtract (2) from (1), we will "eliminate" A - and we get:

B - C = 14 .................................................... (4)

B/C = 5/3.......... → ...... B = 5/3 * C ....... (5)

Now we use (5) in (4) and we get one equation and one unknown

5/3 * C - C = 14

C = 21 (skipped some steps)...................(6)

using (6) in (2) we get .......................................................... A = 40 answer

check

B/C = 5/3 → B = 5/3 * 21 = 35

A + B = 40 + 35 = 75 .........................................checks (1)

A + C = 40 + 21....................................................checks (2)

B/C = 35/21 = 5/3 ............................................. checks (3)

wow, Dr Khan. that was amazing.
A+B=75
A-C=61................................................ This is incorrect. Check the equation again. It should be A + C = 61
B-C =14
Yes, i got it now. i did not see it . not at all. the two A's cancel out and then i'm left with B-C=14
The rest i see clearly now. thanks a lot!
but i want you to check what i did in #21.
ws my substitution right, what was wrong there. seems to me that finding out the individual values of the variables does not work. at least the way i did it. if you can check it please, let me know where i went awry, cos i want to learn it also that way.

 

[lev888]

let me rectify. i think i know i made a mistake here. because i have the value of A and i did not plug it in
Now i will use A to solve the other
A-C=61
solving for C,
A-C=61
75-B-C=61 ADD B TO BOTH SIDES
-B -C+75+B= 61+B
-C+75 = B+ 61 I WILL ADD -75 TO BOTH SIDES

-C +75 -75=B+61 -75
-C=B-14 I WILL DIVIDE BOTH SIDES BY -1 TO GET TID OF THE ENGATICE SIGN ON THE VARIABLE

-C/-1= B-14/-1

C= -B + 14
this is the value of C
SO RECTIFIYING
A=75-B
C=-B+14

now, the value of B

A+B= 75 PLUGGIN' IN VALUES
75-B+B=75
and here is where I am in doubt cos B=0 here.

First of all, you need all 3 equations to solve the problem. So, please add the 3rd one and let's see what happens.
Regarding your solution above, I don't understand what you are doing with the first equation. You wrote "solving for C", but I don't see where you solved the first equation for C. It seems like you are using the second equation. You should only use the first equation. Solve it for C.

 

[Deleted member 4993]

wow, Dr Khan. that was amazing.
A+B=75
A-C=61
B-C =14
Yes, i got it now. i did not see it . not at all. the two A's cancel out and then i'm left with B-C=14
The rest i see clearly now. thanks a lot!
but i want you to check what i did in #21.
ws my substitution right, what was wrong there. seems to me that finding out the individual values of the variables does not work. at least the way i did it. if you can check it please, let me know where i went awry, cos i want to learn it also that way.

You wrote the second equation incorrectly

A + B=75
A+C=61
B - C =14

 

[eddy2017]

First of all, you need all 3 equations to solve the problem. So, please add the 3rd one and let's see what happens.
Regarding your solution above, I don't understand what you are doing with the first equation. You wrote "solving for C", but I don't see where you solved the first equation for C. It seems like you are using the second equation. You should only use the first equation. Solve it for C.

okay. that is what i was doing wrong. i have to use only the first equation. here i go.

There are 75 students in a classes A and B altogether. There are 61 students in classes A and C altogether. The ratio of the number of students in Class B to Class C is 5:3. How many students are there in Class A?

Step 1
i added class A and B and equaled them to 75. Did the same with the classes A and B

A+B=75
A+C =61
one question. how do i set up the ration given as a third equation?. b/c= 5/3.
how do i add this up to the other two?.

 

[lev888]

okay. that is what i was doing wrong. i have to use only the first equation. here i go.

There are 75 students in a classes A and B altogether. There are 61 students in classes A and C altogether. The ratio of the number of students in Class B to Class C is 5:3. How many students are there in Class A?

Step 1
i added class A and B and equaled them to 75. Did the same with the classes A and B

A+B=75
A+C =61
one question. how do i set up the ration given as a third equation?. b/c= 5/3.
how do i add this up to the other two?.

B/C=5/3 is the 3rd equation. What do you mean by "set up the ratio"? The ratio is reflected in the 3rd equation. That's it. Now you need to solve the system:

A+B=75
A+C =61
B/C=5/3

 

[Deleted member 4993]

okay. that is what i was doing wrong. i have to use only the first equation. here i go.

There are 75 students in a classes A and B altogether. There are 61 students in classes A and C altogether. The ratio of the number of students in Class B to Class C is 5:3. How many students are there in Class A?

Step 1
i added class A and B and equaled them to 75. Did the same with the classes A and B

A+B=75 .............................................................(1)
A+C =61.............................................................(2)
one question. how do i set up the ration given as a third equation?. b/c= 5/3........................
how do i add this up to the other two?.

You are not reading carefully the steps I showed you

A+B=75 .............................................................(1)
A+C =61.............................................................(2)
one question. how do i set up the ration given as a third equation?. B/C= 5/3

Read response #22 again and write it down with pencil&paper - answer my friend will be floating on the paper.

After you wrote it (response #22) - line by line - and you still have the question ("how do i set up the ration given as a third equation?") - write back we will work through it.

 

[eddy2017]

B/C=5/3 is the 3rd equation. What do you mean by "set up the ratio"? The ratio is reflected in the 3rd equation. That's it. Now you need to solve the system:

A+B=75
A+C =61
B/C=5/3
''\begin{bmatrix}a+b=75\\ a+c=61\\ \frac{b}{c}=\frac{5}{3}\end{bmatrix}''

 

[pka]

There are 75 students in a classes A and B altogether. There are 61 students in classes A and C altogether. The ratio of the number of students in Class B to Class C is 5:3. How many students are there in Class A?

Let us look at a solution. We are given that: \(A+B=75~\&~A+C=61\)
Subtracting those two we get \(B-C=14\) Now using the ratio we have:
\(\dfrac{B}{C}=\dfrac{5}{3}\text{ or }3B=5C\text{ or }C=\dfrac{3}{5}B\)
Substituting for \(C\) in the line two above we have \(B-\dfrac{3}{5}B=14\)
Multiplying through by \(5\) we get \(5B-3B=70\text{ or } 2B=70\text{ or } B=35\).
From that we get \(A=40~\&~C=21\).

 

[eddy2017]

B/C=5/3 is the 3rd equation. What do you mean by "set up the ratio"? The ratio is reflected in the 3rd equation. That's it. Now you need to solve the system:

A+B=75 ---------------------------1
A+C =61----------------------------2
B/C=5/3----------------------------3

i will isolate a in (1)
a=75-b
i will plug in a in (2)
75-b+c=61-----------------------(4)
b/c=5/3
i will isolate b in(4)
b=61-75-c
b=c+14
so, now i will plug in b=c+14 in (3)
c+14/c=5/3
cross multiplying
(c+14)*3=c*5
3(c+14)=c*5
3c+42=c*5
3c+42-42=c*5-42 i'll simplify
3c=c*5-42 subbin' c*5 from both sides
3c-c*5=c*5-42-c*5 simplifying
-2c=-42 i'll divide by -2
c=21

now solvin' for
b=c+14
b=21+14
b=35

solving for a=75-b
a=75-35
A=40
B=35
C=21
so, now checking
A+B=75
40+35=75

A+C =61
40+21=61

B/C=5/3
35/21=5/3
1.66 repeating = 1.6 repeating

ANSWER :40 STUDENTS IN CLASS A

 

[eddy2017]

B/C=5/3 is the 3rd equation. What do you mean by "set up the ratio"? The ratio is reflected in the 3rd equation. That's it. Now you need to solve the system:

A+B=75 ---------------------------1
A+C =61----------------------------2
B/C=5/3----------------------------3

i will isolate a in (1)
a=75-b
i will plug in a in (2)
75-b+c=61-----------------------(4)
b/c=5/3
i will isolate b in(4)
b=61-75-c
b=c+14
so, now i will plug in b=c+14 in (3)
c+14/c=5/3
cross multiplying
(c+14)*3=c*5
3(c+14)=c*5
3c+42=c*5
3c+42-42=c*5-42 i'll simplify
3c=c*5-42 subbin' c*5 from both sides
3c-c*5=c*5-42-c*5 simplifying
-2c=-42 i'll divide by -2
c=21

now solvin' for
b=c+14
b=21+14
b=35

solving for a=75-b
a=75-35
A=40
B=35
C=21
so, now checking
A+B=75
40+35=75

A+C =61
40+21=61

B/C=5/3
35/21=5/3
1.66 repeating = 1.66 repeating

ANSWER :40 STUDENTS IN CLASS A

 

[eddy2017]

Please, confirm it. i tihnk it is good. if not, i will redo it even if i have to stay here until the wee hours of the morning.

 

[eddy2017]

Let us look at a solution. We are given that: \(A+B=75~\&~A+C=61\)
Subtracting those two we get \(B-C=14\) Now using the ratio we have:
\(\dfrac{B}{C}=\dfrac{5}{3}\text{ or }3B=5C\text{ or }C=\dfrac{3}{5}B\)
Substituting for \(C\) in the line two above we have \(B-\dfrac{3}{5}B=14\)
Multiplying through by \(5\) we get \(5B-3B=70\text{ or } 2B=70\text{ or } B=35\).
From that we get \(A=40~\&~C=21\).

Excellent, pka. I love it. I also wanted to give the substitution method a try. How do you think i did, of course, the way you have shown here and Dr Khan's are way less clunky and its faster.

 

[lev888]

Re post 32:

Here, the second line does not follow from the first:
b=61-75-c
b=c+14

 

[JeffM]

Eddy

I may be confusing the issue, and if so I apologize.

[MATH]a + b = 75.[/MATH] You are told that in the statement of the problem.

[MATH]a + c = 61.[/MATH] You are told that in the statement of the problem.

[MATH]\dfrac{b}{c} = \dfrac{5}{3}.[/MATH] You are told that in the statement of the problem.

Just translate the English into math.

As Subhotosh explained, we SYSTEMATICALLY eliminate variables when we have multiple variables and multiple equations. The third equation already does not have a so that seems like the easiest variable to eliminate. The easiest way to do so in this case (not the only way) is

[MATH](a + b) - (a + c) = 75 - 61 \implies b - c = 14 \implies b = c + 14.[/MATH]
We subtracted to get rid of a. Any questions about that?

Now we have two equations in two unknowns, namely

[MATH]b - c = 14 \text { and } \dfrac{b}{c} = \dfrac{5}{3}.[/MATH]
That last equation is not linear, and I suspect that is what got you lost. But we can fix that issue by multiplying by c.

[MATH]c * \dfrac{b}{c} = c * \dfrac{5}{3} \implies b = c * \dfrac{5}{3} \implies 3b = 5c.[/MATH]
[MATH]b = c + 14 \text { and } 3b = 5c \implies 3(c + 14) = 5c.[/MATH]
We have one equation in one unknown. Solve it and you are done. I am not sure what you want confirmed.

 

[eddy2017]

Re post 32:

Here, the second line does not follow from the first:
b=61-75-c
b=c+14

it shoud have been
i will isolate b for
75-b+c=61 so, (after subtracting from both sides/ simplifying and dividing both sides by -1)
b=c+14

 

[eddy2017]

i understand it now, dear friends. i understand. i was just trying to have lev confirm the solution using the substitution method with a three-equation system. but i get it now. it was hard. never done a ratio problem like that. that is a new for my list of ratio problems and with different approaches.
thanks!.

 

[eddy2017]

Eddy

I may be confusing the issue, and if so I apologize.

[MATH]a + b = 75.[/MATH] You are told that in the statement of the problem.

[MATH]a + c = 61.[/MATH] You are told that in the statement of the problem.

[MATH]\dfrac{b}{c} = \dfrac{5}{3}.[/MATH] You are told that in the statement of the problem.

Just translate the English into math.

As Subhotosh explained, we SYSTEMATICALLY eliminate variables when we have multiple variables and multiple equations. The third equation already does not have a so that seems like the easiest variable to eliminate. The easiest way to do so in this case (not the only way) is

[MATH](a + b) - (a + c) = 75 - 61 \implies b - c = 14 \implies b = c + 14.[/MATH]
We subtracted to get rid of a. Any questions about that?

Now we have two equations in two unknowns, namely

[MATH]b - c = 14 \text { and } \dfrac{b}{c} = \dfrac{5}{3}.[/MATH]
That last equation is not linear, and I suspect that is what got you lost. But we can fix that issue by multiplying by c.

[MATH]c * \dfrac{b}{c} = c * \dfrac{5}{3} \implies b = c * \dfrac{5}{3} \implies 3b = 5c.[/MATH]
[MATH]b = c + 14 \text { and } 3b = 5c \implies 3(c + 14) = 5c.[/MATH]
We have one equation in one unknown. Solve it and you are done. I am not sure what you want confirmed.

yes, that got me lost. you're right. can you please, paste again what you wrote. I do not understand it very well. i want to have all the opinions filed away i n my digital folders but that writing does not let me read it well. if it is not too much to ask, that is.

 

[eddy2017]

Let us look at a solution. We are given that: \(A+B=75~\&~A+C=61\)
Subtracting those two we get \(B-C=14\) Now using the ratio we have:
\(\dfrac{B}{C}=\dfrac{5}{3}\text{ or }3B=5C\text{ or }C=\dfrac{3}{5}B\)
Substituting for \(C\) in the line two above we have \(B-\dfrac{3}{5}B=14\)
Multiplying through by \(5\) we get \(5B-3B=70\text{ or } 2B=70\text{ or } B=35\).
From that we get \(A=40~\&~C=21\).

pka, can i ask you to put what you wrote in clean form, i mean taking away the dfrac {......and stuff like that. i need to file that

 

[JeffM]

yes, that got me lost. you're right. can you please, paste again what you wrote. I do not understand it very well. i want to have all the opinions filed away i n my digital folders but that writing does not let me read it well. if it is not too much to ask, that is.

You can’t take a screen shot?

 

[eddy2017]

You can subtract (and also add) equations like this (LHS = left hand side, RHS = right hand side)...

LHS1 = RHS1
LHS2 = RHS2

Subtracting gives...

LHS1 - LHS2 = RHS1 - RHS2

In your case you have two equations:-

A + B = 75
A + C = 61

When you subtract it can be easier think in the following layout (don't write this in an exam)...

  LHS1     =     RHS1
- LHS2         - RHS2

  (A + B)  =    75
- (A + C)     - 61

  (    B)  =    75
- (    C)     - 61

     B - C =    14

Or doing it on single lines...

(LHS1) - (LHS2) = (RHS1) - (RHS2)

(A + B) - (A + C) = (75) - (61)

A + B - A - C = 14

B - C = 14

pka is direct, and he has a great knowledge of the order that maths should be taught in the educational system. He was just pointing out what stage this is normally taught. I'm sure that he didn't intend any insult. You might have missed the lesson, or the teacher skipped it, or whatever. There's no need to be upset because you can learn it now. As you can see above it's quite easy! Very soon you'll just do these steps in your head.

thank you cubist, a lot and then some. i asked pka to forgive em for such a stupid blowup. it gets to me when i can't see a way into a solution. he accepted my apologies. yes, he is . he is a great mind and a great help here in the forum. his help is inestimable to me.
thanks for your explanation. very interesting. i got it now. totally!