t & w q11

From first equation we get

Vx = Mx + MVy

From 2nd equation we get
x/M-2 = x / (V-2) + 2y (v-2 is in denominator)



x/ (M-2) = { x + 2y (V- 2) } / (V-2)

x (V -2) = (M -2) x + 2y (M-2) (V -2)

then ..what to substitute ?
 
I asked you a question in post 18. Any updates???
Vx = Mx + MVy

From 2nd equation we get
x / (M-2) = x / (V-2) + 2y (v-2 , m-2 is in denominator)



x/ (M-2) = { x + 2y (V- 2) } / (V-2)

x (V -2) = (M -2) x + 2y (M-2) (V -2)

then ..what to substitute ?
 
Vx = Mx + MVy

From 2nd equation we get
x / (M-2) = x / (V-2) + 2y (v-2 , m-2 is in denominator)



x/ (M-2) = { x + 2y (V- 2) } / (V-2)

x (V -2) = (M -2) x + 2y (M-2) (V -2)

then ..what to substitute ?
I asked whether you could solve a system of equations. Why can't you reply?
Assuming you are familiar with the substitution method, can you use the first equation to isolate one variable and the substitute the expression into the second equation? Since the question is about Mini I would isolate V in the first equation to get the final equation with M. Then solve it for M.
 
Since the question is about Mini I would isolate V in the first equation to get the final equation with M. Then solve it for M.
i tried it but nothing came except
Vx = Mx + MVy


V= Mx / (x - M y) --> isolated V


From 2nd equation we get
x / (M-2) = x / (V-2) + 2y (v-2 , m-2 is in denominator)



x/ (M-2) = { x + 2y (V- 2) } / (V-2)

Substituting V=Mx / (x - M y)

x/ (M-2) = { x + 2y (Mx / (x - M y) - 2) } / ( Mx / (x - M y) - 2 )


There are 3 variables M , x , y still left
 
i tried it but nothing came except
Vx = Mx + MVy


V= Mx / (x - M y) --> isolated V


From 2nd equation we get
x / (M-2) = x / (V-2) + 2y (v-2 , m-2 is in denominator)



x/ (M-2) = { x + 2y (V- 2) } / (V-2)

Substituting V=Mx / (x - M y)

x/ (M-2) = { x + 2y (Mx / (x - M y) - 2) } / ( Mx / (x - M y) - 2 )


There are 3 variables M , x , y still left
x and y are constants, not variables. Did you notice that the answers are expressions with x and y? Just try to solve for M.
 
x and y are constants, not variables. Did you notice that the answers are expressions with x and y? Just try to solve for M.
okay i did it .


x/ (M-2) = { x + 2y (V- 2) } / (V-2)

Substituting V=Mx / ( x - M y)


x/ (M-2) = { x + 2y (Mx / (x - M y) - 2) } / ( Mx / (x - M y) - 2 )

x = (Mx - 2x ) + 2yM - 4y

M= ( 3x - 2y M + 4y ) / x

No of questions mini makes in x minutes = x / M = { ( Mx - 2x ) + 2yM - 4y } * x / ( 3x - 2yM + 4y )

Please help
 
x/ (M-2) = { x + 2y (Mx / (x - M y) - 2) } / ( Mx / (x - M y) - 2 )

x = (Mx - 2x ) + 2yM - 4y

M= ( 3x - 2y M + 4y ) / x
Don't get this. You are omitting many intermediate steps. And the last line should express M through other terms that DO NOT include M.
 
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Don't get this. You are omitting many intermediate steps. And the last line should express M through other terms that DO NOT include M.
x/M = x/V + y --(i)
Vx = M x + M V y

V=M x / ( x - M y )


x/M-2 = x / V-2 + 2y ---(ii)

x (V -2) = (M -2) x + 2y (M-2) (V -2)

Substituting V=M x / (x - M y)

x / (M-2) = { x + 2y ( Mx / (x - M y) - 2) } / ( Mx / (x - M y) - 2 )


x = (Mx - 2x ) + 2yM - 4y

M= (3x + 4y) / (x + 2y) [ express M through other terms that DO NOT include M ]


Then ?
 
x/M = x/V + y --(i)
Vx = M x + M V y

V=M x / ( x - M y )


x/M-2 = x / V-2 + 2y ---(ii)

x (V -2) = (M -2) x + 2y (M-2) (V -2)

Substituting V=M x / (x - M y)

x / (M-2) = { x + 2y ( Mx / (x - M y) - 2) } / ( Mx / (x - M y) - 2 )


x = (Mx - 2x ) + 2yM - 4y

M= (3x + 4y) / (x + 2y) [ express M through other terms that DO NOT include M ]


Then ?
Then you have FOUND (calculated) M - and that was your FIND. You are done!!!
 
x/M = x/V + y --(i)
Vx = M x + M V y

V=M x / ( x - M y )


x/M-2 = x / V-2 + 2y ---(ii)

x (V -2) = (M -2) x + 2y (M-2) (V -2)

Substituting V=M x / (x - M y)

x / (M-2) = { x + 2y ( Mx / (x - M y) - 2) } / ( Mx / (x - M y) - 2 )


x = (Mx - 2x ) + 2yM - 4y

M= (3x + 4y) / (x + 2y) [ express M through other terms that DO NOT include M ]


Then ?
I have NO idea what you are doing. You need to slow down and think very carefully about each step.
E.g.
We have
1. V=Mx/(x-My)
2. x (V -2) = (M -2) x + 2y (M-2) (V -2)

Then you write: "Substituting V=M x / (x - M y)"

And you get: x / (M-2) = { x + 2y ( Mx / (x - M y) - 2) } / ( Mx / (x - M y) - 2 )

How???

Please post all intermediate steps if you want help.
 
Q.
Mini and Vinay are quiz masters preparing for a quiz. In 'x' minutes, Mini makes 'y' questions more than Vinay. If it were possible to reduce the time needed by each to make a question by 2 mins , then in 'x' minutes Mini would make '2y' questions more than Vinay. How many questions does Mini make in 'x' minutes?


1] 1/4[ 2 ( x+y) - ( 2 x^2 + 4 x y^2 )^1/2 ]
2] 1/4[ 2(x-y) - ( 2 x^2 + 4 y^2 )^1/2 ]
3] Either option 1 or 2
4] 1/4[ 2(x-y) - ( 2 x^2 - 4 y^2 )^1/2 ]
I dislike this problem because you have to obtain a specific form for the answer. On the other hand, looking at the form of the answers strongly suggests that it will involve the quadratic formula (or, equivalently, completing the square).

But looking even more closely (see bolding), something suggests a typo in the answer; and in fact, I find the problem in various places such as this, where I see that I am right about the typo:

1636503523840.png

I have not stepped into this discussion because it's become a mess of hasty and unchecked and incompletely stated attempts, that I have no desire to unravel. But I did want to point out that reading the problem in its entirety, before doing any work and again from time to time, can be very helpful in choosing a course of action.
 
We have
1. V=Mx/(x-My)
2. x (V -2) = (M -2) x + 2y (M-2) (V -2)

x/M = x/V + y --(i)
V=Mx / (x-My)


x / ( M-2) = x / V-2 + 2y ---(ii)

x/ (M-2) = { x + 2y ( V -2 ) } / ( V - 2 )

Substituting V=M x / (x - M y)

x = (M-2) * { x + 2y ( V -2 ) } / ( V - 2 )

x = (M-2) * { x +2y { M x / (x - M y) - 2 } / ( M x / (x - M y ) - 2 )

x = [ (M-2) * { (x + 2y) / ( x- My) * ( M x - 2 x + 2M y ) } * ( x - M y) ] / ( M x - 2 x + 2M y )


x = (Mx - 2x ) + 2yM - 4y

M= (3x + 4y) / (x + 2y) [ expressing M through other terms that DO NOT include M ]
 
x/M = x/V + y --(i)
V=Mx / (x-My)


x / ( M-2) = x / V-2 + 2y ---(ii)

x/ (M-2) = { x + 2y ( V -2 ) } / ( V - 2 )

Substituting V=M x / (x - M y)

x = (M-2) * { x + 2y ( V -2 ) } / ( V - 2 )

x = (M-2) * { x +2y { M x / (x - M y) - 2 } / ( M x / (x - M y ) - 2 )

x = [ (M-2) * { (x + 2y) / ( x- My) * ( M x - 2 x + 2M y ) } * ( x - M y) ] / ( M x - 2 x + 2M y )


x = (Mx - 2x ) + 2yM - 4y

M= (3x + 4y) / (x + 2y) [ expressing M through other terms that DO NOT include M ]
No of questions mini makes in x minutes = x / M = (Mx - 2x ) + 2yM - 4y / [ (3x + 4y) / (x + 2y) ]

Now , x is (Mx - 2x ) + 2yM - 4y . We cannot have M in x = (Mx - 2x ) + 2yM - 4y

Substituting M = (3x + 4y) / (x + 2y) in (Mx - 2x ) + 2yM - 4y / [ (3x + 4y) / (x + 2y) ]

We get , [ (3x + 4y) * x / (x + 2y) - 2x + 2y * (3x + 4y) / (x + 2y) - 4y ] / [ ( 3x + 4y) / ( x + 2y ) ]

It is getting too complicated .


Answer is A .
 
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x = [ (M-2) * { (x + 2y) / ( x- My) * ( M x - 2 x + 2M y ) } * ( x - M y) ] / ( M x - 2 x + 2M y )


x = (Mx - 2x ) + 2yM - 4y
This is a big leap, and clearly something went wrong, since you should get a quadratic equation. Please show all steps. (I would clear fractions earlier.)

It is getting too complicated .
Yes, this is a complicated problem; I haven't yet obtained their answer, though I did get a quadratic equation and my result had a lot in common with theirs. If I have time, I may go through my work to find an error ... just as you need to learn to do.
 
STARTING FROM beginning.


x/M = x/V + y --(i)

V=Mx / (x-My)

Okay?

THEN ,

x / ( M-2) = x / ( V-2 ) + 2y ---(ii)

x / (M-2) = { x + 2y ( V -2 ) } / ( V - 2 )

Substituting V= M x / (x - M y) IN x = (M-2) * { x + 2y ( V -2 ) } / ( V - 2 )

x = (M-2) * { x +2y { [ M x / (x - M y) ] - 2 } / [ ( M x / (x - M y ) ] - 2 )

Okay up to this?

Then,

x = [ ( M -2 ) { x + 2y { Mx - 2x + 2My } / (x - My) ] / ( Mx - 2x + 2My ) / ( x - My)

x = [ ( M - 2 ) { x + 2y / ( x- My ) * ( Mx - 2x + 2My) } * ( x - My) ] / ( Mx - 2x + 2M y)

Okay upto this ?
 
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