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Solving for a variable Quiz 2

mathematics

 The sum of the ages of Rahul and Sharan is $34$ years. Rahul is elder to Sharan by $6$  years. Find their present ages.

 To solve the question, first assume the ages of both Rahul and Sharan. Then use the conditions given in the question to form two equations. Add the two equations to get your answer.

Let Rahul be $x$ years old and Sharan be $y$ years old. $ \\ $ Sum of their ages is $34$. $ \\ $ So, $x + y = 34{\text{         }}.....{\text{(1)}}$ $ \\ $ W.K.T Rahul is elder to Sharan by $6$  years. So, $x - y = 6{\text{           }}.....{\text{(2)}}$ $ \\ $ Adding equations ${\text{(1)}}$ and ${\text{(2)}}$, we get, $ \\  {\text{2x = 40}} \\ \Rightarrow {\text{x = 20}} \\  $ $ \\ $ Hence Rahul is ${\text{20}}$ years old. $ \\ $ Now putting this value of ${\text{x}}$ in ${\text{(1)}}$, we get $ \\ 20 + y = 34{\text{      }} \\   \Rightarrow y = 14 \\  $ $ \\ $ So, Sharan is $14$ years old.

 Find the value of $\alpha $ if the equation $4x - \alpha y = 6$ has one solution at $\left( {3,3} \right)$.

 To find the solution, equate the point satisfying the equation in the equation. Perform the necessary operations and find the value of $\alpha $.

We are given the equation: $4x - \alpha y = 6$ $ \\ $ Now, $\left( {3,3} \right)$ is one of the solutions of the above equation. $ \\ $ So, put $x = 3,y = 3$ in the above equation. We get: $ \\  4\left( 3 \right) - \alpha \left( 3 \right) = 6 \\ \Rightarrow 6 = 3\alpha  \\ \therefore \alpha  = 2$

  Solve the equations $2x - 3y =  - 24$  and $2x + 4y = 11$. Then find the value of $m$ when $y = mx + 3$.

 To solve the question, subtract the two equations from each other and then find the values of $x$ and $y$. Equate these values in $y = mx + 3$.

Given  $ \\ 2x - 3y =  - 24{\text{      }}....{\text{(1)}} \\ 2x + 4y = 11{\text{         }}....{\text{(2)}} \\ $ Subtract equation ${\text{(2)}}$ from ${\text{(1)}}$, we get $ \\ 2x - 3y - 2x - 4y =  - 24 - 11 \\   \Rightarrow y = 5{\text{ }} \\ $ Substituting value of $y$ in ${\text{(1)}}$, we get $ \\ 2x - 3\left( 5 \right) =  - 24 \\ \Rightarrow 2x =  - 9 \\ \Rightarrow x = \dfrac{{ - 9}}{2} \\ $ So, $x =  - \dfrac{9}{2}$ and $y = 5$. $ \\ $ To find the value of m, $ \\ $  Put these values in the equation $y = mx + 3$. $ \\ $ So, $ 5 = m\left( { - \dfrac{9}{2}} \right) + 3 \\ \Rightarrow 10 =  - 9m + 6 \\ \Rightarrow m =  - \dfrac{4}{9} \\ $

 A lending library charges a fixed rate of Rs $27$ for the first three days and then an additional fee for each day after that. Komal bought a book for Rs $27$ and kept it for $7$ days. If the fixed costs are $x$ and the extra charges are $y$ per day. Find the linear equation, If the ordinate satisfying this equation is $3$, find the abscissa.

 To solve the question use ${\text{Amount paid for fixed days  +  amount paid for additional days  =  total amount paid}}$. After finding the equation, equate the value of ordinate to find the abscissa.

Total amount paid by Komal $ = 27$ $ \\ $ We are given that fixed charges are $x$ and per day additional charges are $y$.  $ \\ $ Out of $27$ Rupees, Rs $x$ were paid for the first three days and Rs $y$ for the next four. $ \\ $ The equation is: Amount paid for fixed days $ + $ Amount paid for additional days $ = $ Total amount paid $ \\ $ So, the required equation is $x + 4y = 27$. $ \\ $ Now, we are given that the ordinate satisfying this equation is $3$. $ \\ $ Let the abscissa be $\alpha $ So, the point is $\left( {\alpha ,3} \right)$. $ \\ $ On equating, we get $ \\ \alpha  + 12 = 27 \\ \Rightarrow \alpha  = 15 \\ <br>&nbsp;$

 In $2$ hours, Raman can row downstream $20$Km and in $2$ hours, he can row upstream $4$Km. Find his speed of rowing in still water and find the speed of current by forming a pair of equations and then solving them.

 To solve the question, first assume Raman’s speed in still water and the speed of the current. Keep in mind the formula ${\text{Distance =  speed}} \times time$ and form two equations with the help of conditions given in the question. Add these equations to get your answer.

Let the speed of the boat be $x$ $km/h$. $ \\ $ Let the speed of the current be $y$ $km/h$. $ \\ $ Now, Raman’s speed during downstream is $(x + y)km/h$ and Raman’s speed during upstream is $(x - y)km/h$ $ \\ $ We know, ${\text{Distance =  speed}} \times time$ $ \\ $ According to the question, $ \\ 2\left( {x + y} \right) = 20 \\ \Rightarrow x + y = 10{\text{            }}.....{\text{(1)}} \\ $ And $ 2\left( {x - y} \right) = 4 \\   \Rightarrow x - y = 2{\text{             }}......{\text{(2)}} \\  $ Now, add equation ${\text{(1)}}$ and ${\text{(2)}}$, we get $ {\text{2x = 12}} \\ \Rightarrow {\text{x = 6}} \\  $ Putting this value in ${\text{(1)}}$, we get, $ \\ x + 6 = 10{\text{  }} \\ \Rightarrow y = 4 \\  $ Hence, Raman’s speed in still water is ${\text{6}}$ $km/h$ and the speed of the current is $4$ $km/h$