Math, asked by sd6357845, 1 month ago

4x4 = 16
nswer any four questions:
a) A variable quantity y is equal to sum of two
quantities, one of which varies directly as x and the other
varies inversely as x. If y = 11 when x = 1 and y = 13 when x = 2, find y when x = 3.​

Answers

Answered by prasadanand76
2

\fontsize{18}{10}\textbf{\textup{The required value of y when x is 3 is 17.}}\fontsize1810The required value of y when x is 3 is 17.

Step-by-step explanation: Given that a variable quantity y is equal to sum of two quantities, one of which varies directly as x and the other varies inversely as x.

Also, y = 11 when x = 1 and y = 13 when x =2.

We are to find the value of y when x = 3.

Let p and q be the two quantities such that y = p + q.

According to the given information, we have

\begin{gathered}p\propto x\\\\\Rightarrow p=kx,~~\textup{where k is a proportionality constant}\end{gathered}

p∝x

⇒p=kx, where k is a proportionality constant

and

\begin{gathered}q\propto \dfrac{1}{x}\\\\\\\Rightarrow q=\dfrac{h}{x},~~\textup{where h is another proportionality constant}.\end{gathered}

q∝

x

1

⇒q=

x

h

, where h is another proportionality constant.

So, we have

y=kx+\dfrac{h}{x}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)y=kx+

x

h

(i)

Given that y = 11 when x = 1.

So, equation (i) implies

\begin{gathered}11=k\times 1+\dfrac{h}{1}\\\\\Rightarrow k+h=11\\\\\Rightarrow k=11-h~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)\end{gathered}

11=k×1+

1

h

⇒k+h=11

⇒k=11−h (ii)

And y = 13 when x = 2. Equation (i) implies

\begin{gathered}13=2k+\dfrac{h}{2}\\\\\Rightarrow 4k+h=26\\\\\Rightarrow 4(11-h)+4=26~~~~~~~~~~~~[\textup{Using equation (ii)}]\\\\\Rightarrow 44-3h=26\\\\\Rightarrow 3h=18\\\\\Rightarrow h=6.\end{gathered}

13=2k+

2

h

⇒4k+h=26

⇒4(11−h)+4=26 [Using equation (ii)]

⇒44−3h=26

⇒3h=18

⇒h=6.

From equation (ii), we get

k=11-6=5.k=11−6=5.

So, from equation (i), we get

y=5x+\dfrac{6}{x}.y=5x+

x

6

.

Therefore, when x = 3, then the value of y is

y=5\times3+\dfrac{6}{3}=15+2=17.y=5×3+

3

6

=15+2=17.

Thus, the required value of y when x is 3 is 17.

Learn more : A variable quantity y is equal to sum of two quantities, one of which varies directly x and the other varies inversely as x. If y = 11 when x = 1 and y = 13 when x = 2, find y when x = 3.

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