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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 two minutes, then in x minutes mini would make 2y questions more than Vinay. How many questions does Mini make in x minute?

1] 1/4[ 2(x+y) - ( 2 x^2 + 4 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 ]

My attempt:

Let Mini and Vinay take M and V minutes to frame a question respectively.

Mini takes M mins to frame 1 question.

Therefore Mini takes $x$ mins to frame $\frac{x}{M}$ questions.

Vinay takes V mins to frame 1 question.

Therefore Vinay takes $x$ mins to frame $\frac{x}{V}$ questions.

ATQ:

$$\frac{x}{M}=\frac{x}{V}+y \tag 1$$

Multiplying the above equation by 2 yields:

$$\frac{2x}{M}=\frac{2x}{V}+2y \tag{1a}$$

Now, in the second scenario:

Mini takes (M-2) mins to frame 1 question.

Therefore Mini takes $x$ mins to frame $\frac{x}{(M-2)}$ questions.

Vinay takes V-2 mins to frame 1 question.

Therefore Vinay takes $x$ mins to frame $\frac{x}{(V-2)} $ questions.

Therefore ATQ:

$$\frac{x}{M-2}=\frac{x}{V-2}+2y \tag 2$$

Equating 2y from $(1)a$ and $2$ yields:

$$\frac{x}{M-2}-\frac{x}{V-2}=\frac{2x}{M}-\frac{2x}{V}$$

$$\frac{1}{M-2}-\frac{1}{V-2}=\frac{2}{M}-\frac{2}{V}$$

$$\frac{V-2-M+2}{(M-2)(V-2)}=\frac{2V-2M}{MV}$$

$$\frac{V-M}{MV-2M-2V+4}=\frac{2V-2M}{MV}$$

$$(V-M)MV=2(V-M)(MV-2M-2V+4)$$

$$MV=2MV-4M-4V+8$$

$$MV-4M-4V+8=0$$

What is the other equation?

If someone can tell us any other shorter method. That would be very helpful.But please do tell the other equation that would solve the problem.Thanks.


Edit: I have done as Rohan has mentioned in the comments.We basically have to find $\frac{x}{M}$.I have arrived at an answer which is close to the actual answer but not the actual answer.Please help me with this. (It would be very helpful if someone poits out my fault and works in the same direction as I had started).I would appreciate if someone does this in a concise way(You don't need to do it my way in this case).I haven't written this in mathjax as it would be very time consuming. Thanks.

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  • $\begingroup$ Why is level of difficulty in the title? $\endgroup$ Dec 15, 2017 at 13:38
  • $\begingroup$ @MichaelBurr Just to let others know it's a little difficult question than the elementary ones. $\endgroup$
    – Soumee
    Dec 15, 2017 at 13:41
  • $\begingroup$ Basically, we need to find $\frac{x}{M}$. So, what you do, is first find $M$ from $(1)$ and then substitute in $(2)$. Solve the quadratic in $M$ and find the required quantity. $\endgroup$
    – user371838
    Dec 16, 2017 at 10:42
  • $\begingroup$ @Rohan I did as you said and obtained: $$\frac{x}{M}=\frac{x}{y}.\frac{2(x+y)\mp \sqrt{4x^2+2y^2}}{4}$$ as the answer. Which is close to the actual answer but not the actual answer. I would show you my actual calculations. (But I need to charge my camera first.) . The way I did took 3 full pages...I am feeling like putting a bounty on this question to do it in an efficient way... $\endgroup$
    – Soumee
    Dec 16, 2017 at 11:55
  • $\begingroup$ @Rohan I have done as you said and I have uploaded the images. $\endgroup$
    – Soumee
    Dec 16, 2017 at 14:08

1 Answer 1

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If in $x$ minutes, Mini makes $a$ questions, then, Mini makes one question in $\frac{x}{a}$ minutes.

In $x$ minutes vinay makes $a-y$ questions, then, Vinay makes a question in $\frac{x}{a-y}$minutes.

Then, if Mini makes a question in $\frac{x}{a}-2$ minutes and Vinay in $\frac{x}{a-y}-2$ minutes, it is given that in $x$ minutes Mini makes $2y$ questions more.

Basically we want to find $a$ in the question. This can be solved by taking: $$\frac{x}{\frac{x}{a}-2}-\frac{x}{\frac{x}{a-y}-2}=2y$$ Solving the required quadratic in $a$, gives us one of the correct options as: $$\boxed{a= \frac{2(x+y)-\sqrt{2x^2+4y^2}}{4}}$$

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