GATE Mechanical 2019 Set 2 | GA | Question: 4

Question

The product of three integers $X$, $Y$ and $Z$ is $192$. $Z$ is equal to $4$ and $P$ is equal to the average of $X$ and $Y$. What is the minimum possible value of $P$?

• A. $6$
• B. $7$
• C. $8$
• D. $9.5$

Comments

 

Given :-   $xyz = 192$

$xy = 48$   {putting z=4}

$AM \geq GM \geq HM$

$(x+y) /2 \geq \sqrt{xy}$

$(x+y) /2 \geq \sqrt{48}$

$(x+y) /2 \geq 6.928$

$P \geq 6.928$

P=7 is answer.

 

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This is called Clarity.

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X * Y = 48, then, if we consider -ve values allowed then,

X = -1 and Y = -48 

X + Y / 2 = -24.5 

 

Since there are no negative options in the question, we have to consider positive values only.

Answer 1

Given

$X*Y*Z = 192$,

$Z=4,$

$P = \left ( \frac{X+Y}{2} \right )$

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Solution

$X*Y*Z = 192$

$\Rightarrow$ $X*Y*4 = 192$

$\Rightarrow$ $X*Y = 48$

Now $48$ should be broken into product of 2 numbers such that they are close to each other because if numbers are close to each other then their average will be minimum..

For eg:- $48$ can be $4*12$, $1*48$, $6*8$, $16*3$ but here we will select $6*8$ because $6$ and $8$ are closest to each other with a difference of $2$.

$\Rightarrow$ $X*Y = 6*8$

$X=6, Y=8$

$P = \left ( \frac{X+Y}{2} \right ) = \left ( \frac{8+6}{2} \right )= 7$

$\therefore$ Option B. $7$ is the correct answer.

Comments

You could have simply use the fact that AM >= GM. And this will will give P > 6.xyz and therefore minimum value is 7.

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just for reference:

in general:

Answer 2

It is given that :- $X*Y*Z= 192$ , $Z=4$. So, $X*Y=48$

Now, $P=\frac{X+Y}{2}=\frac{X+\frac{48}{X}}{2}=\frac{1}{2}\left ( X+\frac{48}{X} \right )$

Now, to find the minimum value of $P,$ Differentiate $P$ with respect to $X$

$\frac{\mathrm{d} P}{\mathrm{d} X} = \frac{1}{2}\left ( 1 -\frac{48}{X^{2}} \right )$

Now, $\frac{\mathrm{d} P}{\mathrm{d} X} = 0$

So, $X= \pm 4\sqrt{3}$

Now, at $X= +4\sqrt{3}$,  $\frac{\mathrm{d}^{2}P }{\mathrm{d} X^{2}} > 0$

So, $P$ will be minimum at $X= +4\sqrt{3}$

So, Minimum value of $P$ will be $\frac{1}{2}\left ( 4\sqrt{3} + \frac{48}{4\sqrt{3}} \right ) = 2\sqrt{3} + \frac{6}{\sqrt{3}} = \frac{12}{\sqrt{3}} = 4\sqrt{3} =6.928….$(An Irrational value)

No options are matching.

For this question, Marks were given to all.

Comments

in question x y z are integer so how x= 4*1.73

i think 7 is correct answer . x and y is 8,6

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Thanks. I did mistake. I marked it as wrong answer and requesting moderators and admins to please remove it as best answer.

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@ ankitgupta.1729 , can’t we use the concept of AM and GM here? i.e. for any two non negative real numbers, their Arithmetic mean is always greater than or equal to their Geometric mean (AM GM reference)

In this case, Arithmetic mean of X and Y is P. From value of z , we get their geometric mean as  $\sqrt{48}$. So according to the inequality, the minimum value of P is  $6.928$ , which is closrst to option B.

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nice :)

you can also write it as an answer.

Answer 3

xyz=192

z=4

xy=192/4

xy=48

p=x+y/2

we all know that least sum of x and y is 14 (x=8 , y=6)

which means p=7

but there comes the problem(p!=7)

in question it is said that x and y are integers

so we can take -1 and -48 whose product is 48 and average is -24.5

and -24.5<7

Comments

@Nikhil gate 2020  nice, you have found the mistake in this question. they should have mentioned $+ve$ integers in this question.

Answer 4

$(X*Y*Z =192$ and $Z=4)$ $\Rightarrow$ $X*Y= 48$

$\rightarrow$Given : $P = \frac{X+Y}{2}$

$\rightarrow$Average of $2$ numbers is minimum when the difference of that $2$ numbers is minimum.

$\rightarrow$We can write $48$ as product of $1*48,2*24,3*16,4*12,6*8$

$\rightarrow$Among all these $6*8$ is the expression which contains the numbers $6$ and $8$ and they have the minimum difference as compared to other expressions.

$\rightarrow$So $X=6$ and $Y=8$

$\Rightarrow$ $P = \frac{X+Y}{2} = \frac{6+8}{2} = 7 $

$\therefore$ Option B. $7$ is the correct answer.