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 ?
 
lev888 said:
I asked you a question in post 18. Any updates???
Click to expand...
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 ?
 
Saumyojit said:
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 ?
Click to expand...
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.
 
lev888 said:
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.
Click to expand...
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
 
Saumyojit said:
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
Click to expand...
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.
 
lev888 said:
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.
Click to expand...
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
 
Saumyojit said:
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
Click to expand...
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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lev888 said:
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.
Click to expand...
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 ?
 
Saumyojit said:
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 ?
Click to expand...
Then you have FOUND (calculated) M - and that was your FIND. You are done!!!
 
Saumyojit said:
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 ?
Click to expand...
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.
 
Saumyojit said:
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 ]
Click to expand...
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.
 
lev888 said:
We have
1. V=Mx/(x-My)
2. x (V -2) = (M -2) x + 2y (M-2) (V -2)
Click to expand...

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 ]
 
Saumyojit said:
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 ]
Click to expand...
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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Saumyojit said:
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
Click to expand...
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.)

Saumyojit said:
It is getting too complicated .
Click to expand...
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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